Question

A 0.5L of a solution is prepared by dissolving 115.5g of NaF in it. The molecular weight of NaF is 41.9g/mol). If I took 0.13L of that solution and diluted it to 0.3, what is the molarity of the resulting solution?

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^2.

When 0.062L of a solution containing 1.69M BaF₂ is diluted to 0.8L, what is the concentration of BaF₂ in the new solution?

Answer to 2 decimal places

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Use standard notation or scientific notation as accepted by blackboard

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^3.

How many milliliters of solution of 2.61 M K₂SO₄ must be used to make 54 mL of a solution that has a concentration of  0.148 M K₂SO₄ ?

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Do not include units when reporting your answer

Use standard notation or scientific notation as accepted by blackboard

e.g. 0.001 or 1E-3, not 1 x 10^^-3

Answer 1

1)

mass(NaF)= 115.5 g

use:

number of mol of NaF,

n = mass of NaF/molar mass of NaF

=(1.155*10^2 g)/(41.9 g/mol)

= 2.757 mol

volume , V = 0.5 L

use:

Molarity,

M = number of mol / volume in L

= 2.757/0.5

= 5.513 M

This is concentration of stock solution.

use dilution formula

M1*V1 = M2*V2

1---> is for stock solution

2---> is for diluted solution

Given:

M1 = 5.513 M

V1 = 0.13 L

V2 = 0.3 L

use:

M1*V1 = M2*V2

M2 = (M1*V1)/V2

M2 = (5.513*0.13)/0.3

M2 = 2.39 M

Answer: 2.39 M

2)

use dilution formula

M1*V1 = M2*V2

1---> is for stock solution

2---> is for diluted solution

Given:

M1 = 1.69 M

V1 = 0.062 mL

V2 = 0.8 mL

use:

M1*V1 = M2*V2

M2 = (M1*V1)/V2

M2 = (1.69*0.062)/0.8

M2 = 0.131 M

Answer: 0.13 M

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