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3 years ago

What is the molality of a solution made by dissolving 15.20 g of i2 in 1.33 mol of diethyl ether, (ch3ch2)2o?

1 answer:

3 0

The molarity of solution made by dissolving 15.20g of i2 in 1.33 mol of diethyl ether (CH3CH2)2O is =0.6M

calculation

molarity =moles of solute/ Kg of the solvent

mole of the solute (i2) = mass /molar mass
the molar mass of i2 = 126.9 x2 = 253.8 g/mol

moles is therefore= 15.2 g/253.8 g/mol = 0.06 moles

calculate the Kg of solvent (CH3CH2)2O
mass = moles x molar mass
molar mass of (CH3CH2)2O= 74 g/mol

mass is therefore = 1.33 moles x 74 g/mol = 98.42 grams
in Kg = 98.42 /1000 =0.09842 Kg

molarity is therefore = 0.06/0.09842 = 0.6 M

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3 years ago

Read 2 more answers

What element has 28 electrons when electrically neutral

muminat

Answer:

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Explanation:

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3 years ago

⦁ Find the concentration of H+, OH-, PH and POH of 0.03 M of magnesium hydroxide which ionizes to the extent of only 1 /3 in aqu

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Answer:

$pH=12.3\\\\pOH=1.7\\$

$[H^+]=5x10^{-13}M$

$[OH^-]=0.02M$

Explanation:

Hello there!

In this case, according to the given ionization of magnesium hydroxide, it is possible for us to set up the following reaction:

$Mg(OH)_2(s)\rightleftharpoons Mg^{2+}(aq)+2OH^-(aq)$

Thus, since the ionization occurs at an extent of 1/3, we can set up the following relationship:

$\frac{1}{3} =\frac{x}{[Mg(OH)_2]}$

Thus, x for this problem is:

$x=\frac{[Mg(OH)_2]}{3}=\frac{0.03M}{3}\\\\x= 0.01M$

Now, according to an ICE table, we have that:

$[OH^-]=2x=2*0.01M=0.02M$

Therefore, we can calculate the H^+, pH and pOH now:

$[H^+]=\frac{1x10^{-14}}{0.02}=5x10^{-13}M$

$pH=-log(5x10^{-13})=12.3\\\\pOH=14-pH=14-12.3=1.7$

Best regards!

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