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3 years ago

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An experiment calls for you to use 100 mL of 0.25 M HNO3 solution. All you have available is a bottle of 3.4 M HNO3. How many mi

lliliters of the 3.4 M HNO3 solution do you need to prepare the desired solution?

1 answer:

uysha [10]3 years ago

3 0

Answer: 7.35mL

Explanation:

C1 = 3.4M

V1 =?

C2 = 0.25M

V2 = 100mL

C1V1 = C2V2

3.4 x V1 = 0.25 x 100

V1 = (0.25 x 100) /3.4

V1 = 7.35mL

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Read 2 more answers

When of alanine are dissolved in of a certain mystery liquid , the freezing point of the solution is less than the freezing poin

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The question is incomplete, the complete question is:

When 177. g of alanine $(C_3H_7NO_2)$ are dissolved in 800.0 g of a certain mystery liquid X, the freezing point of the solution is $5.9^oC$ lower than the freezing point of pure X. On the other hand, when 177.0 g of potassium bromide are dissolved in the same mass of X, the freezing point of the solution is $7.2^oC$ lower than the freezing point of pure X. Calculate the van't Hoff factor for potassium bromide in X.

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<u>Explanation:</u>

Depression in the freezing point is defined as the difference between the freezing point of the pure solvent and the freezing point of the solution.

The expression for the calculation of depression in freezing point is:

$\Delta T_f=i\times K_f\times m$

OR

$\Delta T_f=i\times K_f\times \frac{m_{solute}\times 1000}{M_{solute}\times w_{solvent}\text{(in g)}}$ ......(1)

<u>When alanine is dissolved in mystery liquid X:</u>

$\Delta T_f=5.9^oC$

i = Vant Hoff factor = 1 (for non-electrolytes)

$K_f$ = freezing point depression constant

$m_{solute}$ = Given mass of solute (alanine) = 177. g

$M_{solute}$ = Molar mass of solute (alanine) = 89 g/mol

$w_{solvent}$ = Mass of solvent = 800.0 g

Putting values in equation 1, we get:

$5.9=1\times K_f\times \frac{177\times 1000}{89\times 800}\\\\K_f=\frac{5.9\times 89\times 800}{1\times 177\times 1000}\\\\K_f=2.37^oC/m$

<u>When KBr is dissolved in mystery liquid X:</u>

$\Delta T_f=7.2^oC$

i = Vant Hoff factor = ?

$K_f$ = freezing point depression constant = $2.37^oC/m$

$m_{solute}$ = Given mass of solute (KBr) = 177. g

$M_{solute}$ = Molar mass of solute (KBr) = 119 g/mol

$w_{solvent}$ = Mass of solvent = 800.0 g

Putting values in equation 1, we get:

$7.2=i\times 2.37\times \frac{177\times 1000}{119\times 800}\\\\i=\frac{7.2\times 119\times 800}{2.37\times 177\times 1000}\\\\i=1.63$

Hence, the van't Hoff factor for potassium bromide in X is 1.63

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