Advertisements
Advertisements
प्रश्न
Calculate the mass of sodium sulphate required to prepare its 20% (mass percent) solution in 100g of water?
Advertisements
उत्तर
Let the mass of sodium Sulphate required to be = x g
The mass of solution would be = (x + 100) g x g of solute in (x + 100) g of solution
20% = `"x"/("x" + 100) xx 100`
20x + 2000 = 100x
80x = 2000
x = `200/80`
= 25g
APPEARS IN
संबंधित प्रश्न
If the density of methanol is 0.793 kg L–1, what is its volume needed for making 2.5 L of its 0.25 M solution?
Solve the following problem:
Find out the molar mass of the following compounds:
Mohr’s salt [FeSO4(NH4)2SO4.6H2O]
(At. mass: Cu = 63.5; S = 32; O = 16; H = 1; Na = 23; C = 12; Fe = 56; N = 14)
Solve the following problem:
Write the following number in ordinary decimal form:
3.75 × 10−1
Solve the following problem:
Write the following number in ordinary decimal form:
5.16 × 104
Solve the following problem:
Write the following number in ordinary decimal form:
5.00858585
Sulphuric acid reacts with sodium hydroxide as follows:
\[\ce{H2SO4 + 2NaOH -> Na2SO4 + 2H2O}\]
When 1 L of 0.1 M sulphuric acid solution is allowed to react with 1 L of 0.1 M sodium hydroxide solution, the amount of sodium sulphate formed and its molarity in the solution obtained is:
(i) 0.1 mol L–1
(ii) 7.10 g
(iii) 0.025 mol L–1
(iv) 3.55 g
Which of the following terms are unitless?
(i) Molality
(ii) Molarity
(iii) Mole fraction
(iv) Mass percent
What is the difference between molality and molarity?
An aqueous KCl solution of density 1.20 g mL-1 has a molality of 3.30 mol kg-1. The molarity of the solution in mol L-1 is ______. (Nearest integer)
Find the molality of solution if boiling point increases by 1.75 K and molal elevation constant of solvent is 5K kg mol-1.
