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SSSSS [86.1K]
3 years ago
7

You wish to make a 0.299 M hydroiodic acid solution from a stock solution of 6.00 M hydroiodic acid. How much concentrated acid

must you add to obtain a total volume of 50.0 mL of the dilute solution?
Chemistry
1 answer:
Art [367]3 years ago
5 0

Answer:

V_1=2.49mL

Explanation:

Hello,

In this case, considering that the moles of hydrioiodic acid remain unchanged during the dilution process:

n_{HI}=n_{HI}

One could apply the following equality in terms of molarity:

V_1M_1=V_2M_2

Whereas the subscript 1 accounts for the solution before the dilution and 2 after the dilution, therefore, the required volume of 6.00 M acid is:

V_1=\frac{V_2M_2}{M_1} =\frac{50.0mL*0.299M}{6.00M}=2.49mL

Best regards.

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You dissolve 14 g of Mg(NO3)2 in water and dilute to
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Answer:

0.127M

Explanation:

Molarity of a solution = number of moles (n) ÷ volume (V)

Molar mass of Mg(NO3)2 = 24 + (14 + 16(3)}2

= 24 + {14 + 48}2

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Using the formula, mole = mass/molar mass, to convert mass of Mg(NO3)2 to mole

mole = 14g ÷ 148g/mol

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Volume = 750mL = 750/1000 = 0.75L

Molarity = 0.095mol ÷ 0.75L

Molarity = 0.127M

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