Question

A solution is prepared by dissolving 15.0 g of nh3 in 250.0 g of water. the density of the resulting solution is 0.974 g/ml. the molarity of nh3 in the solution is ________ m.

Answer 1

To solve this problem, first we assume the volume is purely additive. The density of the mixture can then be calculated by the summation of mass fraction of each component divided by its individual density:

1 / ρ mixture = (x NH3 / ρ NH3) + (x H2O / ρ H2O)                        ---> 1

Calculating for mass fraction of NH3:

x NH3 = 15 g / (15 g + 250 g)

x NH3 = 0.0566

Therefore the mass fraction of water is:

x H2O = 1 – x NH3 = 1 – 0.0566

x H2O = 0.9434

Assuming that the density of water is 1 g / mL and substituting the known values back to equation 1:

1 / 0.974 g / mL = [0.0566 / (ρ NH3)] + [0.9434 / (1 g / mL)]

ρ NH3 = 0.680 g / mL

Given the density of NH3, now we can calculate for the volume of NH3:

V NH3 = 15 g / 0.680 g / mL

V NH3 = 22.07 mL

The number of moles NH3 is: (molar mass NH3 is 17.03 g / mol)

n NH3 = 15 g / 17.03 g / mol

n NH3 = 0.881 mol

Therefore the molarity of NH3 in the solution is:

Molarity = 0.881 mol / [(22.07 mL  + 250 mL) * (1L / 1000 mL)

M = 3.238 mol/L = 3.24 M

Answer 2

Answer: The molarity of $NH_3$ in  the solution is 3.24 M.

Explanation:

To calculate the molarity of ammonia in solution, we first need to find the volume of solution and we will calculate the volume of solution by using the density of the solution.

To calculate the volume of an object, we use the formula:

$Density=\frac{\text{Mass of an object}}{\text{Volume of an object}}$

We are given:

Density = $0.974g/mL$

Mass of the solution = (15 + 250)g = 265g

Putting values in above equation, we get:

$0.974g/mL=\frac{265g}{Volume}$

Volume of the solution = 272 mL

Now, to calculate the molarity of the solution, we use the formula:

$Molarity=\frac{w\times 1000}{M\times V\text{(in mL)}}$

where,

w = weight of ammonia = 15 g

M = molar mass of ammonia = 17g/mol

V = Volume of the solution = 272 mL

Putting values in above equation, we get:

$\text{Molarity of ammonia}=\frac{15\times 1000}{17\times 272}\\\\\text{Molarity of ammonia in solution}=3.24M$

Hence, molarity of ammonia in solution is 3.24M