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aivan3 [116]
3 years ago
14

How would you prepare a 1 L solution of 3 M Mgo?

Chemistry
1 answer:
Alborosie3 years ago
3 0

Answer: We need to 120 g of MgO in the beaker and add enough water to reach the 1 L mark.​

Explanation:

Molarity of a solution is defined as the number of moles of solute dissolved per liter of the solution.

Molarity=\frac{n}{V_s}

where,

n = moles of solute

V_s = volume of solution in L

moles of MgO = \frac{\text {given mass}}{\text {Molar mass}}=\frac{xg}{40g/mol}

Now put all the given values in the formula of molality, we get

3M=\frac{x}{40g/mol\times 1L}

x=120g

Therefore, we need to 120 g of MgO in the beaker and add  enough water to reach the 1 L mark.​

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How many grams of NaCl are needed to make 0.800 liter of a 5.00 M solution?
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Answer:

             0.324 g is required to make 5.00 M solution of NaCl in 0.800 L.

Given data:

                 Molarity = 5.00 M

                 Formula Mass = 58.5 g/mol

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              Mass in gram = 5.00 M × 58.5 g/mol × 0.800 L / 1000

               Mass in gram = 0.234 g


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