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3 years ago

What is the concentration of bromide, in ppm, if 115.91 g MgBr2 is dissolved in 1.31 L water.

Chemistry

1 answer:

Ksenya-84 [330]3 years ago

6 0

Answer:- Bromide concentration is 80748 ppm.

Solution:- ppm is milligram per liter. Grams of magnesium bromide are given and we are asked to calculate the ppm concentration of bromide ion, So, let's first calculate the grams of bromide and then divide it by the given volume.

Molar mass of Magnesium bromide = $24.305+2(126.90)=278.105$

$115.91gMgBr_2(\frac{1molMgBr_2}{278.105gMgBr_2})(\frac{2molBr}{1molMgBr_2})(\frac{126.90gBr}{1molBr})$

= 105.78 g Br

Let's convert grams of Br to milligrams:

since, 1 g = 1000 mg

$105.78g(\frac{1000mg}{1g})$

= 105780 mg

concentration of Br = $\frac{105780mg}{1.31L}$

= 80748 mg per liter

Since, 1 ppm = 1 mg per liter

So, the bromide concentration is 80748 ppm.

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Find the enthalpy of neutralization of HCl and NaOH. 137 cm3 of 2.6 mol dm-3 hydrochloric acid was neutralized by 137 cm3 of 2.6

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Answer : The correct option is, (D) 89.39 KJ/mole

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First we have to calculate the moles of HCl and NaOH.

$\text{Moles of HCl}=\text{Concentration of HCl}\times \text{Volume of solution}=2.6mole/L\times 0.137L=0.3562mole$

$\text{Moles of NaOH}=\text{Concentration of NaOH}\times \text{Volume of solution}=2.6mole/L\times 0.137L=0.3562mole$

The balanced chemical reaction will be,

$HCl+NaOH\rightarrow NaCl+H_2O$

From the balanced reaction we conclude that,

As, 1 mole of HCl neutralizes by 1 mole of NaOH

So, 0.3562 mole of HCl neutralizes by 0.3562 mole of NaOH

Thus, the number of neutralized moles = 0.3562 mole

Now we have to calculate the mass of water.

As we know that the density of water is 1 g/ml. So, the mass of water will be:

The volume of water = $137ml+137ml=274ml$

$\text{Mass of water}=\text{Density of water}\times \text{Volume of water}=1g/ml\times 274ml=274g$

Now we have to calculate the heat absorbed during the reaction.

$q=m\times c\times (T_{final}-T_{initial})$

where,

q = heat absorbed = ?

$c$ = specific heat of water = $4.18J/g^oC$

m = mass of water = 274 g

$T_{final}$ = final temperature of water = 325.8 K

$T_{initial}$ = initial temperature of metal = 298 K

Now put all the given values in the above formula, we get:

$q=274g\times 4.18J/g^oC\times (325.8-298)K$

$q=31839.896J=31.84KJ$

Thus, the heat released during the neutralization = -31.84 KJ

Now we have to calculate the enthalpy of neutralization.

$\Delta H=\frac{q}{n}$

where,

$\Delta H$ = enthalpy of neutralization = ?

q = heat released = -31.84 KJ

n = number of moles used in neutralization = 0.3562 mole

$\Delta H=\frac{-31.84KJ}{0.3562mole}=-89.39KJ/mole$

The negative sign indicate the heat released during the reaction.

Therefore, the enthalpy of neutralization is, 89.39 KJ/mole

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