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2 years ago

Calculate the molarity of a solution consisting of 25.0 g of KOH in 3.00 L of solution.

Chemistry

1 answer:

melomori [17]2 years ago

5 0

Answer:

0.15M

Explanation:

The equation for molarity is M= n/L. Where "M" is Molarity, "n" is the number of moles of solute, and "L" is the total liters in solution.

You need to calculate the number of moles from the given grams. The molar mass of KOH is (39.098+ 16 +1.008)= 56.106g. To calculate the mols of KOH, $\frac{25.0g}{1}$ × $\frac{1 mol}{56.106g}$ = 0.44558... mol, you see that the grams unit cancel out leaving you with mol as the unit.

The volume is given in L already so no need to do any conversion. M= $\frac{0.4558mol}{3.00L}$ = 0.1485M ≈ 0.15M

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Answer : The concentration of NaOH is, 0.336 M

Explanation:

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

$n_1M_1V_1=n_2M_2V_2$

where,

$n_1,M_1\text{ and }V_1$ are the n-factor, molarity and volume of acid which is $H_2SO_4$

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

We are given:

$n_1=2\\M_1=0.3564M\\V_1=20.05mL\\n_2=1\\M_2=?\\V_2=43.42mL$

Putting values in above equation, we get:

$2\times 0.3564M\times 20.05mL=1\times M_2\times 43.42mL$

$M_2=0.336M$

Thus, the concentration of NaOH is, 0.336 M

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Which pair of samples contains the same number of oxygen atoms in each compound?0.20 mol Ba(OH)2 and 0.20 mol H2SO40.20 mol Br2O

Whitepunk [10]

Answer:

0.20 mol Br2O and 0.20 mol HBrO have the same number of oxygen atoms

Explanation:

In Ba(OH)2 there are 2 moles of O atoms in every mol of Ba(OH)2.

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In H2SO4 there are 4 moles of O atoms for every mol of H2SO4.

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<em>0.20 mol Br2O and 0.20 mol HBrO</em>

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Number of O atoms in 0.20 moles Br2O = 0.20*1 = 0.20 moles O atom

In HBrO there is 1 mol of O atom in every mol HBrO

Number of O atoms in 0.20 moles HBrO = 0.20 *1 = 0.20 moles O atom

⇒ in 0.20 moles Br2O and 0.20 moles HBrO we <u>have the same</u> number of oxygen atoms

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<em>0.10 mol Na2O and 0.10 mol Na2SO4</em>

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