The amount of sodium carbonate present in 250 cc of M5Na2CO3 of

M5=0.2M;0.2=xM.w×1000250⇒5.3g

The amount of sodium carbonate present in 250 cc of M5Na2CO3 of

M5=0.2M;0.2=xM.w×1000250⇒5.3g

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The amount of sodium carbonate present in 250 cc of $\frac{M}{5} {Na}_{2} {CO}_{3}$ of

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2.65 g

10.6 g

5.3 g

1.06 g

answer is C.

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Detailed Solution

$\frac{M}{5} = 0 .2 M; 0 .2 = \frac{x}{M . w} \times \frac{1000}{250} \Rightarrow 5 .3 g$

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The amount of sodium carbonate present in 250 cc of $\frac{M}{5} {Na}_{2} {CO}_{3}$ of