Question

what is the molarity of a 500ml solution containing 36.5 g of hcl​

Answer 1

Answer:

Step I. Calculation of normality of solution

Mass of HCl=3.65g

Molar mass HCl=1+35.5=36.5u=36.5g mol−1

Basicity of HCl from the formula =1

∴ Equivalent mass HCl=Molar massBascity=(36.5g mol−1)1=(3.65 g)(36.5 g equiv−1)=0.1equiv

Volume of solution in litres =500mL=5001000=0.5L

Normality of solution (N)=No. of equivalent of HClVolume of solution in litres=0.1 equiv.(0.5 L)=0.2 equiv L−1=0.2N

Step II. Calculation of volume of concentrated HCl solution required

Normality of conc. HCl=(N1)=0.2N

Let the volume of conc. HCl needed =V1

Normality of dilute HCl(N2)=0.05N

Volume of dilute HCl(V2)=250mL

Applying normality equation : N1V1≡N2V2

(0.2N)×V1=(0.05N)×(250 mL)

V1=(0.05N)×(250mL)(0.2N)=62.5mL

Step-by-step explanation: