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2 years ago

7

Isoamyl salicylate ( = 208.25 g/mol) has a pleasant aroma and is used in perfumes and soaps. Which of the following combinations

gives a 0.75 m solution of isoamyl salicylate in ethyl alcohol (d = 0.7893 g/mL)?
A. 117.2 g isoamyl salicylate in 950.0 mL of ethyl alcohol
B. 117.2 g isoamyl salicylate in 750.0 mL of ethyl alcohol
C. 117.2 g isoamyl salicylate in 750.0 mL of solution
D. 117.2 g isoamyl salicylate in 592.0 g of ethyl alcohol

2 answers:

Sveta_85 [38]2 years ago

7 0

Molality (m) = moles of solute/ mass of solvent -(i)
moles of solute = 117.2 g/ 208.25 g mol∧-1 => 0.56 moles -(ii)
mass of solvent = density x volume (V) => 0.79 g/ml x V -(iii)
*putting values of m, (ii) & (iii) in (i)
0.75 = 0.56/0.79 V
0.56/(0.75 x 0.79) = V
V = 950 mL

telo118 [61]2 years ago

7 0

<u>Answer:</u> The correct answer is Option C.

<u>Explanation:</u>

We are given:

Molarity of isoamyl salicylate = 0.75 M

This means that 0.75 moles of isoamyl salicylate is present in 1 L or 1000 mL of solution.

To calculate the mass of isoamyl salicylate, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Moles of isoamyl salicylate = 0.75 moles

Molar mass of isoamyl salicylate = 208.25 g/mol

Putting values in above equation, we get:

$0.75mol=\frac{\text{Mass of isoamyl salicylate}}{208.25g/mol}\\\\\text{Mass of isoamyl salicylate}=156.19g$

By applying unitary method, we get:

156.19 grams of isoamyl salicylate is present in 1000 mL of solution.

So, 117.2 grams of isoamyl salicylate will be present in = $\frac{1000mL}{156.19g}\times 117.2g=750mL$ of solution.

Thus, 117.2 grams of isoamyl salicylate is present in 750 mL of solution.

To calculate the mass of solution, we use the equation:

$Density=\frac{Mass}{Volume}$

Density of solution = 0.7893 g/mL

Volume of solution = 1000 mL

Putting values in above equation, we get:

$0.7893g/mL=\frac{\text{Mass of solution}}{1000mL}\\\\\text{Mass of solution}=789.3g$

By applying unitary method, we get:

156.19 grams of isoamyl salicylate is present in 789.3 g of solution.

So, 117.2 grams of isoamyl salicylate will be present in = $\frac{789.3g}{156.19g}\times 117.2g=592g$ of solution.

Thus, 117.2 grams of isoamyl salicylate is present in 592 g of solution.

Hence, the correct answer is Option C.

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