Molarity | MCQs

Explanation

To prepare 250 cm³ (or 250 mL) of a 1 M NaOH solution, you need 10 grams of NaOH.

This is because the molar mass of NaOH is 40 g/mol, and a 1 M solution contains 1 mole of solute per liter of solution.

Therefore, 250 mL (0.25 L) of a 1 M solution would require 0.25 moles of NaOH, which translates to 10 grams (0.25 moles * 40 g/mol). 

Explanation

Molarity is calculated by dividing the number of moles of solute by the volume of solution in liters. 

Therefore, the molarity of pure water is approximately 55.56 moles/L, or approximately 55.5 M. 

Explanation

To calculate formality, we need to know the number of moles of dichloromethanic acid and the volume of the solution in liters.

Number of moles = mass / molecular weight

= 6.447 g / 128.94 g/mol

≈ 0.05 mol

Formality = number of moles / volume (in liters)

= 0.05 mol / 0.5 L

= 0.1 F

Explanation

To find the mass of HCl, we need to calculate the number of moles of HCl in 100 cm^3 (or 0.1 L) of 2M HCl solution, and then multiply it by the molar mass of HCl.

Number of moles = Molarity × Volume (in liters)

= 2 M × 0.1 L

= 0.2 mol

Molar mass of HCl = 36.5 g/mol

Mass of HCl = Number of moles × Molar mass

= 0.2 mol × 36.5 g/mol

= 7.3 g

Explanation

To calculate the molarity, we need to calculate the number of moles of sodium chloride (NaCl) dissolved in 500 cm³ (or 0.5 L) of solution.

Molar mass of NaCl = 58.44 g/mol

Number of moles = mass / molar mass = 40 g / 58.44 g/mol ≈ 0.684 mol

Molarity = number of moles / volume (in liters) = 0.684 mol / 0.5 L ≈ 1.368 M

Rounded to one decimal place, this is approximately 1.4M.