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3 years ago

* The cost of table salt and table sugar is Rs 15 per kg.

Chemistry

1 answer:

Vilka [71]3 years ago

6 0

Cost per mole

Table salt : Rs 0.878

Table sugar : Rs 23.63

<h3>Further explanation</h3>

Given

Cost table salt (NaCl) = 15/kg

Cost table sugar(sucrose-C12H22O11) = 69/kg

Required

cost per mole

Solution

mol of 1 kg Table salt(NaCl ,MW= 58.5 g/mol) :

$\tt mol=\dfrac{1000~g}{58.5}=17.09~mol=Rs~15\rightarrow 1~mol=Rs~0.878$

mol of 1 kg Table sugar(C12H22O11 ,MW= 342 g/mol) :

$\tt mol=\dfrac{1000}{342}=2.92~mol=Rs~69\rightarrow 1~mol=Rs~23.63$

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Answer : The average rate of consumption of $H^+$ during the same time interval is, 2.17 M/s

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$aA+bB\rightarrow cC+dD$

Rate of reaction : It is defined as the change in the concentration of any one of the reactants or products per unit time.

The expression for rate of reaction will be :

$\text{Rate of disappearance of A}=-\frac{1}{a}\frac{d[A]}{dt}$

$\text{Rate of disappearance of B}=-\frac{1}{b}\frac{d[B]}{dt}$

$\text{Rate of formation of C}=+\frac{1}{c}\frac{d[C]}{dt}$

$\text{Rate of formation of D}=+\frac{1}{d}\frac{d[D]}{dt}$

$Rate=-\frac{1}{a}\frac{d[A]}{dt}=-\frac{1}{b}\frac{d[B]}{dt}=+\frac{1}{c}\frac{d[C]}{dt}=+\frac{1}{d}\frac{d[D]}{dt}$

From this we conclude that,

In the rate of reaction, A and B are the reactants and C and D are the products.

a, b, c and d are the stoichiometric coefficient of A, B, C and D respectively.

The negative sign along with the reactant terms is used simply to show that the concentration of the reactant is decreasing and positive sign along with the product terms is used simply to show that the concentration of the product is increasing.

The given rate of reaction is,

$5Br^-(aq)+BrO_3^-(aq)+6H^+(aq)\rightarrow 3Br_2(aq)+3H_2O(l)$

The expression for rate of reaction :

$\text{Rate of disappearance of }Br^-=-\frac{1}{5}\frac{d[Br^-]}{dt}$

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$\text{Rate of formation of }Br_2=+\frac{1}{3}\frac{d[Br_2]}{dt}$

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As we are given:

$\frac{d[Br^-]}{dt}=1.81M/s$

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$-\frac{1}{5}\frac{d[Br^-]}{dt}=-\frac{1}{6}\frac{d[H^+]}{dt}$

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$-\frac{1}{6}\frac{d[H^+]}{dt}=-\frac{1}{5}\frac{d[Br^-]}{dt}$

$\frac{d[H^+]}{dt}=\frac{6}{5}\frac{d[Br^-]}{dt}$

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moles NH3 in excess = 0.0150 - 0.0135 =0.0015
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pOH = pKb + log [NH4+]/ [NH3]= 4.7 + log 0.132/ 0.0147=5.65
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