What volume of water should be added to 50ml of HNO3 having density d = 1.5 and 63%
by weight to have 1M solution.
A.750 ml
B.700ml
C.800 ml
D.1400 ml

To solve this problem, we need to determine the concentration of the given nitric acid solution and then calculate how much water should be added to make it a 1M solution.

First, let's find the molarity of the given HNO3 solution:

Calculate the mass of HNO3 in the solution:
Mass of HNO3 = Volume × Density × % by weight of HNO3
= 50 ml × 1.5 g/ml × 0.63
≈ 47.25 grams

Convert the mass of HNO3 to moles using its molar mass:
Molar mass of HNO3 = 1(1) + 14 + 3(16) = 63 g/mol
Moles of HNO3 = Mass of HNO3 / Molar mass of HNO3
= 47.25 g / 63 g/mol
≈ 0.75 moles

Calculate the molarity (M) of the solution:
Molarity (M) = Moles of solute / Volume of solution (in liters)
= 0.75 moles / (50 ml / 1000)
= 15 M

Now, we want to dilute this solution to make it 1M. Let's denote:
V1 = Volume of the initial solution (50 ml)
C1 = Concentration of the initial solution (15 M)
C2 = Concentration of the final solution (1 M)
V2 = Volume of water to be added (unknown)

We'll use the dilution formula:
C1V1 = C2V2

Substituting the values:
15 M × 50 ml = 1 M × (50 ml + V2)
750 ml = 50 ml + V2
V2 = 750 ml - 50 ml
V2 = 700 ml

So, the correct answer is option B. 700 ml of water should be added to 50 ml of the given HNO3 solution to make it a 1M solution.