Question

A chemist prepares a solution of zinc nitrate Zn(NO3)2 by measuring out 78.3g of zinc nitrate into a 350ml volumetric flask and filling the flask to the mark with water. Calculate the concentration in mol/L of the chemist's zinc nitrate solution. Be sure your answer has the correct number of significant digits.

Answer 1

Answer:

M = 1.18 mol/L

Explanation:

Moles is denoted by given mass divided by the molecular mass ,

Hence ,

$n = \frac{w}{n}$

n = moles ,

w = given mass ,

m = molecular mass .

From the information of the question ,

w = 78.3 g

As we known ,

molecular mass of $Zn(NO_3)_2$ is 189.36 g/mol

m = 189.36 g/mol

moles can be calculated as -

$n = \frac{w}{n}$

$n = \frac{78.3g}{189.36 g/mol}$

n = 0.413 mol

MOLARITY  -

Molarity of a substance , is the number of moles present in a liter of solution .

$M=\frac{n}{V}$

M = molarity ( unit = mol / L or M )

V = volume of solution in liter ( unit = L ),  

n = moles of solute ( unit = mol ),

From the question ,.

V = 350 mL

Since , 1 mL = 10⁻³ L

V = 0.350 L

n = 0.413 mol

Molarity can be calculated as  -

$M=\frac{n}{V}$

$M=\frac{0.413}{0.350}$

M = 1.18 mol/L