The strength of 10−2M Na2CO3 solution in terms of molality will be (density of solution =1.10gmL−1 (Molecular weight of Na2CO3=106gmol

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The strength of $10^{- 2} M {Na}_{2} {CO}_{3}$ solution in terms of molality will be (density of solution $= 1.10 g {mL}^{- 1})$ $(Molecular weight of {Na}_{2} {CO}_{3} = 106 gmol)$

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$9.00 \times 10^{- 3}$

$11.2 \times 10^{- 3}$

$1.5 \times 10^{- 2}$

$5.1 \times 10^{- 3}$

answer is A.

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Detailed Solution

Mass of 1000 ml of the solution = (100 $\times$ 1000) g = 1100 g

Mass of solvent $= (1100 - 10^{- 2} \times 106 = 1098.94 g$

$∴ Molality = \frac{10^{- 2}}{1098.94} \times 1000 \approx 9.00 \times 10^{- 3}$

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