A 0.24 g sample of compound of oxygen and boron was found by analysis to contain 0.096 g of boron and 0.144 g of oxygen. Calculate the percentage composition of the compound by weight.
Mass of Boron = 0.096 g
Маss of Oxygen = 0.144 g
Mass of sample of compound of oxygen and Boron = 0.24 g
% of Boron by weight in the compound = 0.096 x 100 / 0.24 = 40%
% of Oxygen by weight in the compound = 0.144 x 100 / 0.24 = 60%
Pragya tested the solubility of three different substances at different temperatures and collected the data as given below (results are given in the following table, as grams of substance dissolved in 100 grams of water to form a saturated solution).
Substance Dissolved | Tem | per | atu | re | in K |
283 | 293 | 313 | 333 | 353 | |
Potassium nitrate | 21 | 32 | 62 | 106 | 167 |
Sodium chloride | 36 | 36 | 36 | 37 | 37 |
Potassium chloride | 35 | 35 | 40 | 46 | 54 |
Ammonium chloride | 24 | 37 | 41 | 55 | 66 |
(a) What mass of potassium nitrate would be needed to produce a saturated solution of potassium nitrate in 50 grams of water at 313 K?
(b) Pragya makes a saturated solution of potassium chloride in water at 353 K and leaves the solution to cool at room temperature. What would she observe as the solution cools? Explain.
(c) Find the solubility of each salt at 293 K. Which salt has the highest solubility at this temperature?
(d) What is the effect of change of temperature on the solubility of a salt?
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Welcome to the NCERT Solutions for Class 9 Science - Chapter . This page offers a step-by-step solution to the specific question from Excercise 6 , Question 1: A 0.24 g sample of compound of oxygen and boron was found by analysis to contain 0.096 g of boron an....
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