Calculate the quantity of sodium carbonates(anhydrous ) required to prepare ` 250 ml ` solution

Options

A. ` 2.65 grams `
B. ` 4.95 grams `
C. ` 6.25grams `
D. None of these

Answer

A

Explanation

We know that
Molarity `= (W)/(MxxV)`
where ,
W = Mass of `Na_(2)CO_(3)` in gram
M = Molecular mass of `Na_(2)CO_(3)` in gram
= 106
V = Volume of solution in litre
`=(250)/(1000)= 0.25`
Molarity `= (1)/(10)`
Hence, `(1)/(10)=(W)/(106xx0.25)`
or `W =(106xx0.25)/(10)= 2.65` grams