CHEMISTRY HIGH SCHOOL

in a school’s laboratory, students require 50.0 mL of 2.50 M H2SO4 for an experiment, but the only available stock solution of the acid has a concentration of 18.0 M. What volume of the stock solution would they use to make the required solution? Use mc018-1.jpg.

Answers

Answer 1

Answer:

Ans: Volume of stock H2SO4 required = 6.94 ml

Given:

Concentration of stock H2SO4 solution M1 = 18.0 M

Concentration of the final H2SO4 solution needed M2 = 2.50 M

Final volume of H2SO4 needed, V2 = 50.0 ml

To determine:

Volume of stock needed, V1

Explanation:

Use the dilution relation:

$M1V1 = M2V2\n\nV1 = (M2V2)/(M1) \n\nV1 = (2.50 M * 50.0 ml)/(18.0 M) = 6.94 ml$

Answer 2

Answer:

Hello!

In a school’s laboratory, students require 50.0 mL of 2.50 M H2SO4 for an experiment, but the only available stock solution of the acid has a concentration of 18.0 M. What volume of the stock solution would they use to make the required solution?

We have the following data:

M1 (initial molarity) = 2.50 M (or mol/L)

V1 (initial volume) = 50.0 mL → 0.05 L

M2 (final molarity) = 18.0 M (or mol/L)

V2 (final volume) = ? (in mL)

Let's use the formula of dilution and molarity, so we have:

$M_(1) * V_(1) = M_(2) * V_(2)$

$2.50 * 0.05 = 18.0 * V_(2)$

$0.125 = 18.0\:V_2$

$18.0\:V_2 = 0.125$

$V_2 = (0.125)/(18.0)$

$V_2 \approx 0.00694\:L \to \boxed{\boxed{V_2 \approx 6.94\:mL}}\:\:\:\:\:\:\bf\green{\checkmark}$

Answer:

The volume is approximately 6.94 mL

_______________________________

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Answers

Answer:

12 mol of water

Explanation:

$6CO_2 + 6H_2O \longrightarrow C_6H_(12)O_6 + 6O_2$

First we must find the moles of CO2.

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By stoichiometry we know that for every 6 moles of carbon dioxide 6 moles of water are needed, now if we have 12 moles of carbon dioxide how many moles of water will be needed

We apply a simple rule of three

$6 molCO_2 \longrightarrow 6 mol H_2O\n 12 mol CO_2 \longrightarrow x\nx= ((12).(6))/(6)\n x= 12 mol H_2O$

Given:
6CO2 + 6H2O -> C6H12O6 + 6O2
molar mass of carbon dioxide (CO2) is 44.01 g/mol
molar mass of water (H2O) is 18.01 g/mol
528 g of CO2

Required:
moles of water

Solution:

528 g of CO2/ (44.01 g/mol CO2) = 12 moles CO2

12 moles CO2 (6 moles H2O/ 6 moles CO2) = 12 moles H2O

12 moles H2O (18.01 g/mol H2O) = 216.12 grams H2O

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Answers

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Answers

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