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3 years ago

How many milliliters of an aqueous solution of 0.222 M manganese(II) iodide is needed to obtain 9.57 grams of the salt?

Chemistry

1 answer:

Vaselesa [24]3 years ago

8 0

Answer:

139 mL

Explanation:

Given data:

Volume of solution required = ?

Molarity of solution = 0.222 M

Mass of salt = 9.57 g

Solution:

Number of moles of manganese iodide:

Number of moles = mass/molar mass

Number of moles = 9.57 g/ 308.75 g/mol

Number of moles = 0.031 mol

Volume required:

Molarity = number of moles of solute / volume in L

0.222 M = 0.031 mol / V

V = 0.031 mol / 0.222 mol/L

V = 0.139 L

Volume in mL:

0.139 L ×1000 mL / 1 L

139 mL

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A chemistry student weighs out of sulfurous acid , a diprotic acid, into a volumetric flask and dilutes to the mark with distill

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The question is incomplete, here is the complete question:

A chemistry student weighs out 0.104 g of sulfurous acid, a diprotic acid, into a 250.0 mL volumetric flask and dilutes to the mark with distilled water. He plans to titrate the acid with 0.0700 M NaOH solution. Calculate the volume of NaOH solution the student will need to add to reach the final equivalence point. Be sure your answer has the correct number of significant digits.

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Given mass of sulfurous acid = 0.104 g

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To calculate the volume of base, we use the equation given by neutralization reaction:

$n_1M_1V_1=n_2M_2V_2$

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$n_2,M_2\text{ and }V_2$ are the n-factor, molarity and volume of base which is NaOH.

We are given:

$n_1=2\\M_1=5.07\times 10^{-3}M\\V_1=250mL\\n_2=1\\M_2=0.0700M\\V_2=?mL$

Putting values in above equation, we get:

$2\times 5.07\times 10^{-3}\times 250.0=1\times 0.0700\times V_2\\\\V_2=\frac{2\times 5.07\times 10^{-3}\times 250}{1\times 0.0700}=36.2mL$

Hence, the volume of NaOH needed is 36.2 mL

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3 years ago

How many milliliters of an aqueous solution of 0.222 M manganese(II) iodide is needed to obtain 9.57 grams of the salt?​

How many milliliters of an aqueous solution of 0.222 M manganese(II) iodide is needed to obtain 9.57 grams of the salt?