6-2 g of a sample containing Na2CO3 amp- non-volatile inert impurity on gentle heating loses 5

Dear Student, Weight #160;loss #160;is #160;due #160;to #160;conversion #160;of #160;NaHCO3 #160;to #160;Na2CO32NaHCO3 #160;= #160;Na2C ...

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Standard XII

Chemistry

Enthalpy of Solution

6.2 g of a sa...

Question

6.2 g of a sample containing Na2CO3 & non-volatile inert impurity on gentle heating loses 5% of its weight due to reaction 2NaHCO3 = Na2CO3 + H2O + CO2. Residue is dissolved in water an formed 100 cc solution & its 10 cc portion requires 7.5 cc of 0.2 M aqueous solution of BaCl2 for complete precipitation of carbonates. Determine weight of Na2CO3 in the original sample.

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Solution

Dear Student,

$Weight loss is due to conversion of {NaHCO}_{3} to {Na}_{2} {CO}_{3} 2 {NaHCO}_{3} = {Na}_{2} {CO}_{3} + H_{2} O + {CO}_{2} 2 mol 1 mol 2 mol of {NaHCO}_{3} corresponds to 2 \times 84 = 168 g 1 mol of {Na}_{2} {CO}_{3} = 106 g So, loss in weight = 168 - 106 = 62 g or, we can say that 31 g weight is lost per mole of {NaHCO}_{3} and, 5 % of 6.2 g of mixture is lost i . e . 0.3 g of weight loss occurs from \frac{0.3}{31} mol of {NaHCO}_{3} producing \frac{0.3}{62} mol of {Na}_{2} CO 3 Total moles of carbonate = Total moles of {BaCl}_{2} So, M_{1} \times 10 = 0.2 \times 7.5 or M_{1} = 0.15 M i . e . Molarity of carbonate is 0.15 M So, moles of carbonate in 100 ml solution = \frac{0.15 \times 100}{1000} = 0.015 mol Now moles of carbonate in original sample = 0.015 - \frac{0.3}{62} = 0.01 Mass of {Na}_{2} {CO}_{3} in original sample = 0.01 \times 106 = 1.06 g$

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6.2

g of a sample containing

N a_{2} C O_{3}, N a H C O_{3}

and non-volatile inert impurity on gentle heating loses

5 %

of its weight due to reaction:

2 N a H C O_{3} \to N a_{2} C O_{3} + H_{2} O + C O_{2}

. Residue is dissolved in water and formed

100 m L

solution and its

10 m L

portion requires

7.5 m L

0.2 M

aqueous solution of

B a C l_{2}

for complete precipitation of carbonates. Determine weight (in g) of

N a_{2} C O_{3}

in the original sample.

$6.2$ g of a sample containing $N a_{2} C O_{3}, N a H C O_{3}$ and non-volatile inert impurity on gentle heating loses $5 %$ of its weight due to reaction as shown below:

$2 N a H C O_{3} \to N a_{2} C O_{3} + H_{2} O + C O_{2}$

Residue is dissolved in water and formed $100$ mL solution and its $10$ mL portion requires $7.5$ mL of $0.2 M$ aqueous solution of $B a C l_{2}$ for complete precipitation of carbonates. Determine the weight (in g) of $N a_{2} C O_{3}$ in the original sample (as nearest integer).