Home
Class 11
CHEMISTRY
The specific gravity of a salt solution ...

The specific gravity of a salt solution is 1.025. If `V mL` of water is added to `1 L` of this solution to make its density `10.2 g mL^(-1)`, what value of `V` in `mL` approxmately?

Text Solution

Verified by Experts

`V_("New") = (V + 1000) mL`
Number of moles of salt remains same.
`:. (V + 1000) xx 10.2 = 1.025 xx 1000`
`V = 4.9 mL = 5 mL`
Promotional Banner

Topper's Solved these Questions

  • SOME BASIC CONCEPTS AND MOLE CONCEPT

    CENGAGE CHEMISTRY|Exercise Archives Single Correct|17 Videos

  • SOME BASIC CONCEPTS AND MOLE CONCEPT

    CENGAGE CHEMISTRY|Exercise Archives Integer|4 Videos

  • SOME BASIC CONCEPTS AND MOLE CONCEPT

    CENGAGE CHEMISTRY|Exercise Exercises Assertion-Reasoning|10 Videos

  • S-BLOCK GROUP 2 - ALKALINE EARTH METALS

    CENGAGE CHEMISTRY|Exercise Ex 5.1 Objective|2 Videos

  • STATES OF MATTER

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

Similar Questions

Explore conceptually related problems

200 ml of water is added of 500 ml of 0.2 M solution. What is the molarity of this diluted solution?

The density of a 2.05 molar solution of acetic acid in water is 1.82g/ml. The molality of the solution is

If 11 g of oxalic acid are dissolved in 500 mL of solution (density =.1.1 g mL^(-1) ), what is the mass % of oxalic acid in solution ?

If 11 g of oxalic acid are dissolved in 500 mL of solution (density = 1.1 g mL^(-1) ), what is the mass % of oxalic acid in solution ?

What will be the Molarity of solution when water is added to 10g CaCO_(3) to make 100mL of solution ? .