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3 years ago

Calculate the molarity of sodium chloride in a half-normal saline solution (0.45% NaCl). The molar mass of NaCl is

Chemistry

1 answer:

Rus_ich [418]3 years ago

3 0

Answer:

0.077 M

Explanation:

Data Given :

The concentration of half normal (NaCl) saline = 0.45g / 100 g

So,

Volume of Solution = 100 g = 100 mL

Volume of Solution in Liter = 100 mL / 1000

Volume of Solution = 0.1 L

molar mass of NaCl = 58.44 g/mol

Molarity:

Molarity is the representation of the solution. It is amount of solute in moles per liter of solution and represented by M

Formula used for Molarity

M = moles of solute / Liter of solution . . . . . . . . . . (1)

Now to find number of moles of Nacl

no. of moles of NaCl = mass of NaCl / molar mass

no. of moles of NaCl = 0.45g / 58.44 g/mol

no. of moles of NaCl = 0.0077 g

Put values in the eq (1)

M = moles of solute / Liter of solution . . . . . . . . . . (1)

M = 0.0077 g / 0.1 L

M = 0.077 M

So the molarity of half-normal saline solution (0.45% NaCl) = 0.077 M

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What is the concentration of chloride ion when 16.8ml of a 0.554 M solution of barium chloride is combined with 16.8ml of a 0.59

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Answer:

The concentration of chloride ion is 2.887 M.

Explanation:

Here we have a combination BaCl₂ and AlCl₃

1 mole BaCl₂ can produce 2 moles chloride ions as follows;

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The decomposition of NH4HS is endothermic: NH4HS(s)⇌NH3(g)+H2S(g) Part A Which change to an equilibrium mixture of this reaction

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Explanation:

According to Le Chatelier's principle, any disturbance caused in an equilibrium reaction will shift the equilibrium in a direction that will oppose the change.

As the given reaction is as follows.

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(a) When increase the temperature of the reactants or system then equilibrium will shift in forward direction where there is less temperature. It is possible for an endothermic reaction.

Thus, formation of $H_{2}S$ will increase.

(b) When we decrease the volume (at constant temperature) of given reaction mixture then it implies that there will be increase in pressure of the system. So, equilibrium will shift in a direction where there will be decrease in composition of gaseous phase. That is, in the backward direction reaction will shift.

Hence, formation of $H_{2}S$ will decrease with decrease in volume.

When we increase the mount of $NH_{4}HS$ then equilibrium will shift in the direction of decrease in concentration that is, in the forward direction.

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