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two miscible liquids were mixed to the point of saturation. it was foun…

Question

two miscible liquids were mixed to the point of saturation. it was found that there was 25.00 ml of liquid a in every 350.00 ml of the solution.

  1. what is the % v/v?

1 ml a
% v/v = ---------------- 100% = 2 %
3 ml soln

  1. how many ml of liquid a are in 25.00 ml of the solution ? (dont use % here, use actual numbers)

4 ml a
(5 ml soln)(----------------) = 6 ml a
7 ml soln
a. 36.00 b. 5.00 c. 31.00 d. 150.0 e. 12 f. 350.00
g. 100.0 h. 2.4 x 10^{-2} i. 1.2 x 10^{-3} j. 1.786 k. 7.143 l. 25.00
m. 7.500 n. 125.0 o. 500.0 p. 15.00 q. 185.00 r. 444.44
s. 2.00 x 10^{3} t. 20.67 u. 0.500 v. 499.5 w. 15 x. 33.33
y. 1.00 x 10^{5} z. 10.0 aa. 1.00 bb. 10^{6} cc. 6.00
dd. 6.00 x 10^{-5} ee. 6.00 x 10^{6} ff. 20.0 gg. 200 hh. 1.20 x 10^{6}
ii. 60 jj. 0.100 kk. 4.00 ll. 10^{3} mm. 0.0667 nn. 66.7

Explanation:

Step1: Calculate \(\%v/v\)

The formula for \(\%v/v=\frac{\text{Volume of solute (A)}}{\text{Volume of solution}}\times100\%\). Here, volume of solute (A) \(V_A = 25.00\) mL and volume of solution \(V_{soln}=350.00\) mL.
So, \(\%v/v=\frac{25.00}{350.00}\times100\%\)

$$=\frac{2500}{350.00}\%=\frac{250}{35}\%=\frac{50}{7}\%\approx7.143\%$$

Step2: Calculate volume of A in \(25.00\) mL solution

Let \(x\) be the volume of A in \(25.00\) mL solution. Using the ratio from \(\%v/v\), we have \(\frac{V_A}{V_{soln}}=\frac{x}{25.00}\). Since \(\frac{V_A}{V_{soln}}=\frac{25.00}{350.00}\), then \(x=(25.00)\times\frac{25.00}{350.00}\)

$$x=\frac{625}{350}\text{mL}\approx1.786\text{mL}$$

Answer:

  1. \(1 = 25.00\), \(2 = 7.143\), \(3 = 350.00\)
  2. \(4 = 25.00\), \(5 = 25.00\), \(6 = 1.786\), \(7 = 350.00\)