Formation of polyethene from calcium carbide takes place as follows : CaC 2 + H 2 O → Ca ( OH ) 2 + C 2 H 2 ; C 2 H 2 + H 2 → C 2 H 4 ; n C 2 H 4 → - CH 2 - CH 2 - n Determine the amount of polyethylene possibly obtainable from 64 . 0 kg CaC 2 can be:

The balanced chemical equations of process are as follows, CaC 2 + 2 H 2 O → Ca ( OH ) 2 + C 2 H 2 C 2 H 2 + H 2 → C 2 H 4 ; n C 2 H 4 → - CH 2 - CH 2 - n From the balanced chemical equations: mol of CaC 2 = mol of C 2 H 2 = mol of C 2 H 4 Moles of C 2 H 4 = mass of CaC 2 molar mass of CaC 2 = 64 × 10 3 64 = 10 3 mol Moles of polyethene = mol of C 2 H 4 n mass of polyethene molar mass of polyethene = mol of C 2 H 4 n mass of polyethene 28 n = 10 3 n Mass of polyethene = 28 × 10 3 g = 28 kg

Formation of polyethene from calcium carbide takes place as follows : CaC 2 + H 2 O → Ca ( OH ) 2 + C 2 H 2 ; C 2 H 2 + H 2 → C 2 H 4 ; n C 2 H 4 → - CH 2 - CH 2 - n Determine the amount of polyethylene possibly obtainable from 64 . 0 kg CaC 2 can be:

The balanced chemical equations of process are as follows, CaC 2 + 2 H 2 O → Ca ( OH ) 2 + C 2 H 2 C 2 H 2 + H 2 → C 2 H 4 ; n C 2 H 4 → - CH 2 - CH 2 - n From the balanced chemical equations: mol of CaC 2 = mol of C 2 H 2 = mol of C 2 H 4 Moles of C 2 H 4 = mass of CaC 2 molar mass of CaC 2 = 64 × 10 3 64 = 10 3 mol Moles of polyethene = mol of C 2 H 4 n mass of polyethene molar mass of polyethene = mol of C 2 H 4 n mass of polyethene 28 n = 10 3 n Mass of polyethene = 28 × 10 3 g = 28 kg

E M B I B E

Practice Book for KVPY Aptitude Test - Stream SA Chemistry>Chapter 1 - Mole Concept - 1>EXERCISE>Q 45

MEDIUM

KVPY Aptitude Test - Stream SA

IMPORTANT

Earn 100

Formation of polyethene from calcium carbide takes place as follows :
${CaC}_{2} + H_{2} O \to Ca (OH)_{2} + C_{2} H_{2};$ $C_{2} H_{2} + H_{2} \to C_{2} H_{4};$

$n (C_{2} H_{4}) \to {(- {CH}_{2} - {CH}_{2} -)}_{n}$
Determine the amount of polyethylene possibly obtainable from $64.0 kg {CaC}_{2}$ can be:

Important Questions on Mole Concept - 1

EASY

KVPY Aptitude Test - Stream SA

IMPORTANT

Practice Book for KVPY Aptitude Test - Stream SA Chemistry>Chapter 1 - Mole Concept - 1>EXERCISE>Q 46

The molality of a sulphuric acid solution is

0.2

. Calculate the total weight of the solution having

1000 gm

of solvent.

EASY

KVPY Aptitude Test - Stream SA

IMPORTANT

Practice Book for KVPY Aptitude Test - Stream SA Chemistry>Chapter 1 - Mole Concept - 1>EXERCISE>Q 47

When

170 g {NH}_{3}

decomposes how many grams of

N_{2}

&

H_{2}

is produced.

EASY

KVPY Aptitude Test - Stream SA

IMPORTANT

Practice Book for KVPY Aptitude Test - Stream SA Chemistry>Chapter 1 - Mole Concept - 1>EXERCISE>Q 48

340 g {NH}_{3} (M = 17)

when decompose how many litres of nitrogen gas is produced at

STP

.

MEDIUM

KVPY Aptitude Test - Stream SA

IMPORTANT

Practice Book for KVPY Aptitude Test - Stream SA Chemistry>Chapter 1 - Mole Concept - 1>EXERCISE>Q 49

4

mole of

{MgCO}_{3}

is reacted with

6

moles of

HCl

solution. Find the volume of

{CO}_{2}

gas produced at

STP,

the reaction is :

{MgCO}_{3} + 2 HCl \to {MgCl}_{2} + {CO}_{2} + H_{2} O

EASY

KVPY Aptitude Test - Stream SA

IMPORTANT

Practice Book for KVPY Aptitude Test - Stream SA Chemistry>Chapter 1 - Mole Concept - 1>EXERCISE>Q 50

117 gm NaCl

is dissolved in

500 mL

aqueous solution. Find the molarity of the solution.

EASY

KVPY Aptitude Test - Stream SA

IMPORTANT

Practice Book for KVPY Aptitude Test - Stream SA Chemistry>Chapter 1 - Mole Concept - 1>KVPY PROBLEMS (PREVIOUS YEARS)>Q 1

Mass of a liquid is weighed correct to three decimal place and its volume is measured correct to one decimal place. The density of the liquid calculated from the above data will be correct to:

EASY

KVPY Aptitude Test - Stream SA

IMPORTANT

Practice Book for KVPY Aptitude Test - Stream SA Chemistry>Chapter 1 - Mole Concept - 1>KVPY PROBLEMS (PREVIOUS YEARS)>Q 2