What is the molarity of the solution produced when 145 g NaCl is dissolved in sufficient water to prepare 2.75 L of solution? *

a. 0.0190
b. 52.7
c. 399
d. 0.902

Given:

We know the molar mass,

Now,

The molar mass of NaCl will be:

=

=

The amount of moles will be:

=

=

=

hence,

The molarity will be:

→

By substituting the values,

Thus the above approach i.e., "option d" is correct.

Learn more about molarity here:

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Option D) 0.902M

Explanation:

Sodium chloride has a chemical formula of NaCl

Given that,

Amount of moles of NaCl (n) = ?

Mass of NaCl in grams = 145g

For molar mass of NaCl, use the molar masses:

Sodium, Na = 23g;

Chlorine,Cl = 35.5g

NaOH = (23g + 35.5g)

= 58.5g/mol

Since, amount of moles = mass in grams / molar mass

n = 145g / 58.5g/mol

n = 2.48 mole

Amount of moles of NaCl (n) = 2.48 mole

Volume of NaCl solution (v) = 2.75L

Concentration of NaCl solution (c) = ?

Since concentration (c) is obtained by dividing the amount of solute dissolved by the volume of solvent, hence

c = n / v

c = 2.48 mole / 2.75L

c = 0.9018M

Approximately, c = 0.902M

[c is the concentration in moles per litres which is also known as molarity]

Thus, the molarity of the solution is 0.902M

the answer is alpha particles hope i you

part 1: the half-life of the element is 22 years.

part 2: it will take 132.02 year for the sample to decay from 308.0 g to 4.8125 g.

explanation:

part 1: there are two ways to determine the half life time.

first way: the half life time can be determined directly from its definition as it is the time required for the sample to decay to its half concentration.

from the data the initial concentration was 45.0 g and decayed to its half value (22.5 g) in 22.0 years.

second way: the radio active decay is considered first order reaction so we can use the laws of first order reactions.

k = (1/t) ln (a/a-x), where k is the rate constant of the reaction,

t is the time of the reaction

a is the initial concentration of the reactant

a-x is the remaining concentration at time (t)

a = 45.0 g the initial concentration at t = 0 year.

at t = 22 year, a-x = 22.5 g

then k = (1/22 year) ln (45.0 g/ 22.5 g) = 0.0315 year-1.

the half-life time (t1/2) = ln 2/ k = 0.693/ k =0.693/ 0.0315 = 22 years.

part 2:

using the same law k = (1/t) ln (a/a-x),

using the given data; a = 308.0 g and a-x = 4.8125 g and the calculated k from the part 1 (k = 0.0315 year-1)

t = (1/k) ln (a/a-x) = (1/0.0315) ln (308.0/ 4.8125) = 132.02 years