Home
Class 11
CHEMISTRY
What is would be the molality of a solut...

What is would be the molality of a solution obtained by mixing equal volumes of 30% by weight `H_(2) SO_(4) (d = 1.218 g mL^(-1))` and 70% by weight `H_(2) SO_(4) (d = 1.610 g mL^(-1))`? If the resulting solution has density `1.425 g mL^(-1)`, calculate its molality.

Text Solution

AI Generated Solution

To solve the problem of finding the molality of a solution obtained by mixing equal volumes of 30% by weight H₂SO₄ and 70% by weight H₂SO₄, we will follow these steps: ### Step 1: Calculate the mass of each solution 1. **For the 30% H₂SO₄ solution:** - Density (d₁) = 1.218 g/mL - Volume (V) = V mL - Mass (m₁) = d₁ × V = 1.218 g/mL × V mL = 1.218V g ...
Promotional Banner

Topper's Solved these Questions

  • SOME BASIC CONCEPTS AND MOLE CONCEPT

    CENGAGE CHEMISTRY ENGLISH|Exercise Ex 1.1 Objective Questions|6 Videos

  • SOME BASIC CONCEPTS AND MOLE CONCEPT

    CENGAGE CHEMISTRY ENGLISH|Exercise Ex 1.1 Objective Questions (Single Correct)|3 Videos

  • S-BLOCK GROUP 2 - ALKALINE EARTH METALS

    CENGAGE CHEMISTRY ENGLISH|Exercise Ex 5.1 Objective|2 Videos

  • STATES OF MATTER

    CENGAGE CHEMISTRY ENGLISH|Exercise Exercises (Ture False)|25 Videos

Similar Questions

Explore conceptually related problems

Calculate the molality and molarity of a solution made by mixing equal volumes of 30% by weight of H_(2)SO_(4) (density =1.20g//ml ) and 70% by weight of H_(2)SO_(4) (density =1.60g//mL )

The molarity of 98% by wt. H_(2)SO_(4) (d = 1.8 g/ml) is

Calculate the pH of solution obtained by mixing 10 ml of 0.1 M HCl and 40 ml of 0.2 M H_(2)SO_(4)

What volume of 90% alcohol by weight (d = 0.8 g mL^(-1)) must be used to prepared 80 mL of 10% alcohol by weight (d = 0.9 g mL^(-1))

What volume of 90% alcohol by weight (d = 0.8 g mL^(-1)) must be used to prepared 80 mL of 10% alcohol by weight (d = 0.9 g mL^(-1))

The molality of 1M NaNO_(3) solution is (d=1.25 g/ml)

10 mL of 1 NHCl is mixed with 20 mL of 1 MH_(2)SO_(4) and 30 mL of 1M NaOH. The resultant solution has:

If 100 mL of 1 N H_(2)SO_(4) is mixed with 100 mL of 1 M NaOH solution. The resulting solution will be

The resulting solution obtained by mixing 100 ml 0.1 m H_2SO_4 and 50 ml 0.4 M NaOH will be

If 100 mL "of" 1 M H_(2) SO_(4) solution is mixed with 100 mL of 98% (W//W) of H_(2)SO_(4) solution (d = 0.1 g mL^(-1)) , then