Question

a) How many mL of a 12.0M solution of HI are needed to make up 250.0 mL of a 0.500 M solution of HI?

b) How many grams of NaNO2 are needed to make of 500 mL of a 1.30 M solution of sodium nitrite?

Please include explanation.

Answer 1

a)

use dilution formula

M1*V1 = M2*V2

Here:

M1 is molarity of solution before dilution

M2 is molarity of solution after dilution

V1 is volume of solution before dilution

V2 is volume of solution after dilution

we have:

M1 = 12.0 M

M2 = 0.500 M

V2 = 250.0 mL

we have below equation to be used:

M1*V1 = M2*V2

V1 = (M2 * V2) / M1

V1 = (0.500*250.0)/12.0

V1 = 10.4 mL

Answer: 10.4 mL

b)

volume , V = 500 mL= 0.500 L

we have below equation to be used:

number of mol,

n = Molarity * Volume

= 1.3*0.500

= 0.650 mol

Molar mass of NaNO2 = 1*MM(Na) + 1*MM(N) + 2*MM(O)

= 1*22.99 + 1*14.01 + 2*16.0

= 69 g/mol

we have below equation to be used:

mass of NaNO2,

m = number of mol * molar mass

= 0.650 mol * 69 g/mol

= 44.8 g

Answer: 44.8 g