Federal regulations set an upper limit of 50 parts per million ( ppm ) of NH 3 in the air in a work environment (that is, 50 mL NH 3 per 10 6 mL of air). The density of NH 3 ( g ) at room temperature is 0 . 771 g L - 1 . Air from a manufacturing operation was drawn through a solution containing 100 mL of 0 . 0105 M HCl . The NH 3 reacts with HCl as follows: NH 3 ( aqueous ) + HCl ( aqueous ) → NH 4 Cl ( aqueous ) After drawing air through the acid solution for 10 minutes at a rate of 10 litres min - 1 , the acid was titrated. The remaining acid required 13 . 1 mL of 0 . 0588 M NaOH to reach the equivalence point. a How many grams of NH 3 were drawn into the acid solution? b How many ppm of NH 3 were in the air? c Is this manufacturer in compliance with regulations?

a 50 ppm = 50 mL NH 3 per 10 6 mL of air Density of NH 3 at room temperature = 0 . 771 g litre - 1 NH 3 + HCl → NH 4 Cl 100 mL 0 . 0105 M In 10 minutes, the volume of air passed = 10 litre min - 1 × 10 min = 100 litre air ∴ meq of NH 3 present in 100 litre air = meq of HCl reacted = 100 × 0 . 105 - 13 . 1 × 0 . 058 mmole of NH 3 in 100 litr e air = 0 . 2797 weight 17 in 10 5 mL air = 0 . 2797 weight of NH 3 drawn to acid solution = 0 . 2797 × 17 = 4 . 755 g b weight of NH 3 in 10 5 mL air = 0 . 2797 × 10 × 17 = 47 . 55 g ∴ Volume of NH 3 present = weight density = 47 . 55 0 . 771 = 61 . 6 mL ∴ ppm of NH 3 = 61 . 6 ppm c No, because maximum allowed ppm is 50 ppm

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$0.9546 g$ of a Rochelle salt, ${NaKC}_{4} H_{4} O_{6} \cdot 4 H_{2} O,$ on ignition, gave ${NaKCO}_{3},$ which was treated with $41.72 mL$ of $0.1307 N H_{2} {SO}_{4}$ . The unreacted $H_{2} {SO}_{4}$ was then neutralised by $1.91 mL$ of $0.1297 N$ $NaOH$ . Find the percentage purity of the Rochelle salt in the sample.

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Important Questions on Volumetric Calculation

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Modern Approach to Chemical Calculations>Chapter 7 - Volumetric Calculation>PROBLEMS>Q 50

A mixture of ${KMnO}_{4}$ and $K_{2} {Cr}_{2} O_{7}$ weighing $0.24 g$ on being treated with $KI$ in acid solution liberates just sufficient $I_{2}$ to react with $60 mL$ of $0.1 N {Na}_{2} S_{2} O_{3}$ solution. Calculate the percentage of $Cr$ and $Mn$ in the mixture.

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JEE Main

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Modern Approach to Chemical Calculations>Chapter 7 - Volumetric Calculation>PROBLEMS>Q 51

Federal regulations set an upper limit of $50$ parts per million $(ppm)$ of ${NH}_{3}$ in the air in a work environment (that is, $50 mL {NH}_{3}$ per $10^{6} mL$ of air). The density of ${NH}_{3} (g)$ at room temperature is $0.771 g L^{- 1}$ . Air from a manufacturing operation was drawn through a solution containing $100 mL$ of $0.0105 M$ $HCl$ . The ${NH}_{3}$ reacts with $HCl$ as follows:

${NH}_{3} (aqueous) + HCl (aqueous) \to {NH}_{4} Cl (aqueous)$

After drawing air through the acid solution for $10$ minutes at a rate of $10 litres \min^{- 1}$ , the acid was titrated. The remaining acid required $13.1 mL$ of $0.0588 M NaOH$ to reach the equivalence point.

$(a)$ How many grams of ${NH}_{3}$ were drawn into the acid solution?

$(b)$ How many $ppm$ of ${NH}_{3}$ were in the air?

$(c)$ Is this manufacturer in compliance with regulations?

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JEE Main

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Modern Approach to Chemical Calculations>Chapter 7 - Volumetric Calculation>Objective Problems>Q 2

In the reaction,

2 S_{2} O_{3}^{2 -} + I_{2} \to S_{4} O_{6}^{2 -} + 2 I^{-}

, the equivalent weight of

{Na}_{2} S_{2} O_{3}

is equal to its

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JEE Main

IMPORTANT

Modern Approach to Chemical Calculations>Chapter 7 - Volumetric Calculation>Objective Problems>Q 3

In the reaction,

VO + {Fe}_{2} O_{3} \to FeO + V_{2} O_{5}

, the equivalent weight of

V_{2} O_{5}

is equal to its

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JEE Main

IMPORTANT

Modern Approach to Chemical Calculations>Chapter 7 - Volumetric Calculation>Objective Problems>Q 4

In the reaction,

{Na}_{2} {CO}_{3} + HCl \to {NaHCO}_{3} + NaCl

, the equivalent weight of

{Na}_{2} {CO}_{3}

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Modern Approach to Chemical Calculations>Chapter 7 - Volumetric Calculation>Objective Problems>Q 5

The equivalent weight of iodine in

I_{2} + 2 S_{2} O_{3}^{2 -} \to 2 I^{-} + S_{4} O_{6}^{2 -}

, is equal to its

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JEE Main

IMPORTANT

Modern Approach to Chemical Calculations>Chapter 7 - Volumetric Calculation>Objective Problems>Q 6

The equivalent weight of

K_{2} {Cr}_{2} O_{7}

as an oxidising agent in acid medium is

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JEE Main

IMPORTANT

Modern Approach to Chemical Calculations>Chapter 7 - Volumetric Calculation>Objective Problems>Q 7

In alkaline condition, ${KMnO}_{4}$ reacts as follows:

$2 {KMnO}_{4} + 2 KOH \to 2 K_{2} {MnO}_{4} + H_{2} O + O$ ,

The equivalent weight for ${KMnO}_{4}$ is

0.9546 g of a Rochelle salt, NaKC4H4O6·4H2O, on ignition, gave NaKCO3, which was treated with 41.72 mL of 0.1307 N H2SO4. The unreacted H2SO4 was then neutralised by 1.91 mL of 0.1297 N NaOH. Find the percentage purity of the Rochelle salt in the sample.

Important Questions on Volumetric Calculation

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