The amount of disorder in a system is measured by it's:Activation energy
Equilibrium position
Entropy
Keq

The reactant that is totally consumed during a reaction, or the limiting reagent, decides when the process comes to an end. A limiting reagent is the reactant that is used up first in a reaction. Therefore, option A is correct.

What is a limiting reagent?

A reactant that is completely consumed at the end of a chemical reaction is known as the limiting reagent. Since the reaction cannot proceed without this reagent, the amount of product that can be produced is constrained.

The limiting reactant is the one that is consumed first and sets a limit on the quantity of product(s) that can be produced. Calculate how many moles of each reactant are present and contrast this ratio with the mole ratio of the reactants in the balanced chemical equation to get the limiting reactant.

The terms limiting reagent and limiting agent may also be used to describe the limiting reactant.

Thus, option A is correct.

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Answer: A.) The reactant that is used up first in a reaction

The molarity of the solution is 5.018 mol/L and molality of the solution 5.9846 mol/kg.

Assume 100 g of solution the salt concentration- 35 % and water will be 65 %.    

So,  

mass of salt = 35 g  

mass of water = 65 g  

Number of moles,  

So,    

moles of salt = 35 g / 90 (g/mol) = 0.389 mole  

moles of water = 65 / 18 = 3.6111 mole  

Volume of solution,

volume = 100 g / ( 1.29 g/ml)  

volume of solution = 77.52 ml = 0.07752 L  

a)

molality = 0.389 mol / 0.065 kg  

molality = 5.9846 mol/kg  

b)  

molarity = moles of salt / volume of solution (L)

molarity = 0.389 mol / 0.07752 L  

molarity = 5.018 mol/L  

c)    

 mole fraction of salt = moles of salt / total moles  

mole fraction of salt = 0.389 mol / ( 0.389 + 3.6111)  

mole fraction of salt = 0.09725

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Answer:

a) 5.9846 mol/kg

b) 5.018 mol/L

c)0.09725

Explanation:

consider 100 g of solution

now

since the salt is 35 % , water will be 65 %

now

mass of salt = 35 g

mass of water = 65 g

we know that

moles = mass / molar mass

so

moles of salt = 35 g / 90 (g/mol) = 0.389 mol

moles of water = 65 / 18 = 3.6111 mol

now

volume of solution = mass of solution / density of solution

volume of solution = 100 g / ( 1.29 g/ml)

volume of solution = 77.52 ml

volume of solution = 0.07752 L

a)molality = moles of salt / mass of water (kg)

molality = 0.389 mol / 0.065 kg

molality = 5.9846 mol/kg

b)

molarity = moles of salt / volume of solution (L)

molarity = 0.389 mol / 0.07752 L

molarity = 5.018 mol/L

c)

now

total moles in the solution = moles of salt + moles of water

total moles in the solution = 0.389 + 3.6111

total moles in the solution = 4 mol

now

mole fraction of salt = moles of salt / total moles

mole fraction of salt = 0.389 mol / 4 mol

mole fraction of salt = 0.09725

Answer:  4) 40 ml

Solution :

According to the neutralization law,

where,

= molarity of NaOH solution = 0.30 M

= volume of NaOH solution = ?

= molarity of HCl solution = 0.80 M

= volume of HCl solution = 15 ml

Now put all the given values in the above law, we get the volume of NaOH solution.

Therefore, the volume of 0.30 M NaOH required to neutralize 15 ml of 0.80 M HCl is 40 ml.

Final answer:

The volume of 0.30 M NaOH(aq) needed to neutralize 15.0 milliliters of 0.80 M HCl(aq) is 40 milliliters.

Explanation:

The volume of 0.30 M NaOH(aq) needed to neutralize 15.0 milliliters of 0.80 M HCl(aq) can be determined using the concept of stoichiometry and the balanced equation for the reaction between NaOH and HCl. The balanced equation is:

NaOH(aq) + HCl(aq) → NaCl(aq) + H2O(l)

From the balanced equation, we can see that 1 mole of NaOH reacts with 1 mole of HCl. Therefore, the number of moles of HCl in 15.0 milliliters of 0.80 M HCl solution can be calculated as:

Moles of HCl = Volume (in liters) × Molarity = 0.015 liters × 0.80 moles/liter = 0.012 moles

Since 1 mole of NaOH reacts with 1 mole of HCl, we need 0.012 moles of NaOH. The volume of 0.30 M NaOH solution can be determined by rearranging the equation:

Volume (in liters) = Moles / Molarity = 0.012 moles / 0.30 moles/liter = 0.04 liters = 40 milliliters

Therefore, the volume of 0.30 M NaOH(aq) needed to completely neutralize 15.0 milliliters of 0.80 M HCl(aq) is 40 milliliters.

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