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Byju's Answer

Standard IX

Chemistry

Ways of Representing the Concentration of a Solution

A sample of s...

Question

A sample of sodium carbonate contain sodium sulphate also. $1.5 g$ of the sample is dissolved in water and volume raised to $250 m L$ . $25 m L$ of this solution requires $20 m L$ of $\frac{N}{10} H_{2} S O_{4}$ solution for neutralisation. Calculate percentage of sodium carbonate in the sample.

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Solution

Only $N a_{2} C O_{3}$ will react with $H_{2} S O_{4}$ .
Applying $N_{1} V_{1} (N a_{2} C O_{3}) = N_{2} V_{2} (H_{2} S O_{4})$
$N_{1} \times 25 = 20 \times \frac{1}{10}$
$N_{1} = \frac{20}{25 \times 10} = 0.08$
Eq. mass of $N a_{2} C O_{3} = \frac{M o l . m a s s}{2} = \frac{106}{2} = 53$
Mass of $N a_{2} C O_{3}$ present in $250 m L 0.08 N$ solution
$= \frac{N \times E \times V}{1000} = \frac{0.08 \times 53 \times 250}{1000} = 1.06 g$ $
Percentage of $N a_{2} C O_{3}$ in the mixture $= \frac{1.06}{1.50} \times 100 = 70.67$ .

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A sample of sodium carbonate contain sodium sulphate also. 1.5 g of the sample is dissolved in water and volume raised to 250 mL. 25 mL of this solution requires 20 mL of N10H2SO4 solution for neutralisation. Calculate percentage of sodium carbonate in the sample.

Only Na2CO3 will react with H2SO4.Applying N1V1(Na2CO3)=N2V2(H2SO4)N1×25=20×110N1=2025×10=0.08Eq. mass of Na2CO3=Mol.mass2=1062=53Mass of Na2CO3 present in 250 mL 0.08 N solution=N×E×V1000=0.08×53×2501000=1.06 g$Percentage of Na2CO3 in the mixture =1.061.50×100=70.67.

A sample of sodium carbonate contain sodium sulphate also. $1.5 g$ of the sample is dissolved in water and volume raised to $250 m L$ . $25 m L$ of this solution requires $20 m L$ of $\frac{N}{10} H_{2} S O_{4}$ solution for neutralisation. Calculate percentage of sodium carbonate in the sample.