99 Percentile Qs Bank for JEE MainChemistrySome Basic Concepts of Chemistry
Calculate the molality of 1 ~L solution of 93 % H ₂ SO ₄ by w / V [d_ H ₂ SO ₄ =1.84 ~g / cC ]
Options
- A3.71
- B8.5
- C12.4
- D1.042
Correct answer
D. 1.042
Step-by-step solution
Given, [d_ H ₂ SO ₄ =1.84 ~g / cc ] aligned & Molality (m)= Number of moles of solute Weight of solvent in kg . & aligned & Weight =93 ~g & Molecular weight =98 ~g & Solute = weight mol 1 weight of solvent & d= M V m=d V & m=1.84 gmL ⁻¹ 1000 ~mL =1840 ~g & Weight of H ₂ SO ₄= 93 100 1000=930 ~g aligned aligned So, weight of solvent aligned & =1840-930 ~g & =910 ~g =0.91 ~kg & = 93 ~g 98 ~g 1 0.91 ~kg & =1.042 ~mol / kg aligned