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4 years ago

What is the molarity of a solution of 14.0 g NH4Br in enough H2O to make 150 mL of solution?

2 answers:

8 0

The molarity of a solution equals to the mole number of the solute/the volume of the solution. For NH4Br, we know that the mole mass is 98. So the molarity is (14/98) mol /0.15 L=0.95 mol/L.

aniked [119]4 years ago

4 0

Explanation:

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

Mathematically, Molarity = $\frac{no. of moles}{volume of solution in liter}$

Since it is given that the molarity of a solution of 14.0 g and volume is 150 mL or 0.15 L.

Whereas number of moles = $\frac{mass}{molar mass}$

So, molar mass of $NH_{4}Br$ is 97.94 g/mol.

Thus, number of moles = $\frac{14.0 g}{97.94 g/mol}$

= 0.142 mol

Therefore, calculate the molarity as follows.

Molarity = $\frac{no. of moles}{volume of solution in liter}$

= $\frac{0.142 mole}{0.15 L}$

= 0.946 mol/L

Hence, we can conclude that molarity of the solution is 0.946 mol/L.

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An aqueous solution of calcium hydroxide is standardized by titration with a 0.120 M solution of hydrobromic acid. If 16.5 mL of

Artist 52 [7]

<u>Answer:</u> The molarity of calcium hydroxide in the solution is 0.1 M

<u>Explanation:</u>

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 $HBr$

$n_2,M_2\text{ and }V_2$ are the n-factor, molarity and volume of base which is $Ca(OH)_2$

We are given:

$n_1=1\\M_1=0.120M\\V_1=27.5mL\\n_2=2\\M_2=?M\\V_2=16.5mL$

Putting values in above equation, we get:

$1\times 0.120\times 27.5=2\times M_2\times 16.5\\\\M_2=0.1M$

Hence, the molarity of $Ca(OH)_2$ in the solution is 0.1 M.

7 0

3 years ago

Part III.The two reactions involved in quantitatively determining the amount of iodate in solution are: IO3-(aq) 5 I-(aq) 6 H (a

slega [8]

Answer:

$\large \boxed{\math{\dfrac{\text{6 mol thiosulfate}}{\text{1 mol iodate}}}}$

Explanation:

The I₂ is the common substance in the two equations.

(1) IO₃⁻ + 5I⁻ + 6H⁺ ⟶ 3I₂ + 3H₂O

{2) I₂ + 2S₂O₃²⁻ ⟶ 2I⁻ + S₄O₆²⁻

From Equation (1), the molar ratio of iodate to iodine is

$\dfrac{\text{I}_{2}}{\text{IO}_{3}^{-}} = \dfrac{3}{1}$

From Equation (2), the molar ratio of iodine to thiosulfate is

$\dfrac{\text{S$_{2}$O}_{3}^{2-}}{\text{I}_{2}} = \dfrac{2}{1}$

Combining the two ratios, we get

$\text{Stoichiometric factor} = \dfrac{\text{S$_{2}$O}_{3}^{2-}}{\text{IO}_{3}^{-}} = \dfrac{\text{S$_{2}$O}_{3}^{2-}}{\text{I}_{2}} \times \dfrac{\text{I}_{2}}{\text{IO}_{3}^{-}} = \dfrac{2}{1} \times \dfrac{3}{1} = \mathbf{\dfrac{6}{1}}\\\\\text{The stoichiometric factor is $\large \boxed{\mathbf{\dfrac{\text{6 mol thiosulfate}}{\text{1 mol iodate}}}}$}$