Question

The amount of anhydrous Na2CO3 present in
250 ml of 0.25 M solution is
(1) 6.225 g
(3) 6.0 g
(2) 66.25 g
(4) 6.625 g
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Answer 1

Answer:

option C is correct

Explanation:

M= mole of solute/volume of solution ×1000

0.25 =no. of mole of solute/ 250× 1000

no. of mole of solute = o.625

molar mass of Na2CO3= 106g

mole=given mass /molar mass

given mass=mole × molar mass

given mass = o.625 ×106

given mass = 66.25

Answer 2

Answer:

M=no. of moles of solute /vol. of solution in lt*1000

no. of moles of solute =wt/molar mass

let wt =X, molar mass =106

no of moles of solute =x/106

molarity =0.25

vol. of solution =250

M=x*1000/Molar mass *vol. of solution

0.25=x*1000/106*250

x=1000/0.25*250*106

x= 8

amount of anhydrous =8