250 mL of a Na2CO3 solution contains 2.65 g of Na2CO3 ⋅ 10 mL of this solution is added to x mL of water to obtain 0.001 M Na2CO3 solution. The value of x is (molecular weight of Na2CO3 = 106 )

Question

250 mL of a Na2CO3 solution contains 2.65 g of Na2CO3 ⋅ 10 mL of this solution is added to x mL of water to obtain 0.001 M Na2CO3 solution. The value of x is (molecular weight of Na2CO3 = 106 ) (A) 1000 mL (B) 990 mL (C) 9990 mL (D) 90 mL

Tardigrade Admin · Accepted Answer

Given, volume of Na 2 CO 3 solution =250 mL and, mass of Na 2 CO 3 solution =2.65 g molecular mass of Na 2 CO 3=106 g ∴ Eq wt of Na 2 CO 3=(106/2)=53 ∵ w =(E N V/1000) ∴ 2.65 =(53 × N × 250/1000) N =0.2 ∵ 10 mL of this solution is added to x mL of water to obtain 0.001 M Na 2 CO 3 solution, therefore volume of solution taken, V1=10 mL normality of solution, N1=0.2 N volume of solution after addition of water V2=10+x Normality of solution, N2=0.001 M =0.002 N (∵. For .Na 2 CO 3, N=2 M ) From, N1 V1=N2 V2 10 × 0.2=(10+x) × 0.002 x=990 mL

Q. 250mL of a Na2​CO3​ solution contains 2.65g of Na2​CO3​⋅10mL of this solution is added to xmL of water to obtain 0.001M Na2​CO3​ solution. The value of x is (molecular weight of Na2​CO3​=106 )

1000 mL

990 mL

9990 mL

90 mL

Q. $250 m L$ of a $N a_{2} C O_{3}$ solution contains $2.65 g$ of $N a_{2} C O_{3} \cdot 10 m L$ of this solution is added to $x m L$ of water to obtain $0.001 M$ $N a_{2} C O_{3}$ solution. The value of $x$ is (molecular weight of $N a_{2} C O_{3} = 106$ )