1.25 g of a sample of Na_ 2 CO_ 3 and Na_ 2 SO_ 4 is dissolved in 250 ml solution. 25 ml of this solution neutralises 20 ml of 0.1 N H_ 2 SO_ 4 . The \% of Na_ 2 CO_ 3 in this sample is

Correct answer: A. 84.8 \% Let the amount of Na_ 2 CO_ 3 present in the mixture be xg. Na_ 2 SO_ 4 will not react with H_ 2 SO_ 4 . Then (x)/(53)=(20 × 0.1 × 10)/(1000) x=1.06 g Percentage of Na_ 2 CO_ 3 =(1.06 × 100)/(1.25)=84.8 \%

BITSAT 2016

Exam: BITSAT
Year: 2016
Subject: Chemistry
Chapter: Some Basic Concepts of Chemistry

BITSAT 2016 · Chemistry · Some Basic Concepts of Chemistry

1.25 g of a sample of Na_ 2 CO_ 3 and Na_ 2 SO_ 4 is dissolved in...

1.25 \mathrm{ g}

of a sample of

\mathrm{Na}_{2} \mathrm{CO}_{3}

and

\mathrm{Na}_{2} \mathrm{SO}_{4}

is dissolved in

250 \mathrm{ml}

solution.

25 \mathrm{ml}

of this solution neutralises

20 \mathrm{ml}

0.1 \mathrm{ N} \mathrm{H}_{2} \mathrm{SO}_{4}

. The

\%

\mathrm{Na}_{2} \mathrm{CO}_{3}

in this sample is

Options

A $84.8 \%$
B $8.48 \%$
C $15.2 \%$
D $42.4 \%$

Correct answer

A. $84.8 \%$

Step-by-step solution

Let the amount of

\mathrm{Na}_{2} \mathrm{CO}_{3}

present in the mixture be

\mathrm{xg}. \mathrm{Na}_{2} \mathrm{SO}_{4}

will not react with

\mathrm{H}_{2} \mathrm{SO}_{4}.

Then

\frac{x}{53}=\frac{20 \times 0.1 \times 10}{1000} \therefore x=1.06 g\therefore

Percentage of

\mathrm{Na}_{2} \mathrm{CO}_{3}=\frac{1.06 \times 100}{1.25}=84.8 \%

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