Question

3.0 molal sodium hydroxide (naoh) solution has a density of 1.110g/ml. the molarity of this solution is​

Answer 1

Answer:

Answer:

Molarity = 2.973 M

Steps:

Molality of the given NaOH soln: 3 molal

Density of the solution = 1.11 g/ml

According to Molality formula,

$\boxed{ Molality = \dfrac{ \text{No. of moles of solute}}{\text{Mass of Solvent in kg}}}$

Hence 3 molal NaOH solution would contain 3 moles of NaOH in 1 kg of water.

Hence Mass of NaOH = No of moles of NaOH × Molar Mass of NaOH

Mass of NaOH = 3 × ( 23 + 16 + 1 ) ⇒ 3 × 40 = 120g

Hence Mass of Solution = Mass of NaOH + Mass of Water

→ Mass of Solution = 120g + 1000g = 1120g

Now we know that,

⇒ Density of Soln. = Mass of Soln. / Volume of Soln.

Hence, Volume of Soln. is given as:

⇒ Volume of Soln. = Mass of Soln. / Density

⇒ Volume of Soln. = 1120g / 1.11 g/ml

⇒ Volume of Soln. = 1009 ml (approx.)

Hence Molarity of the solution can be given as:

$\implies Molarity = \dfrac{\text{No of moles of solute}}{\text{Volume of Soln. in liters}}$

Substituting in the formula we get:

$\implies \boxed{Molarity = \dfrac{3}{1.009} = 2.973\: M}$

Hence the molarity of the solution is 2.973 M.

Answer 2

$Answer:</p><p>Given \\ \\ A \: body \: of \: mass \: 500 \: kg \: \\ started from \: rest \: and \\ moves \\ with \: a \\ speed \: of \: 10 \: m/s \\ in \: 5 s \: \\ To \: Find \\ Force \\ Solution \\ We \: know that \\ \rm \longrightarrow F=m \times a \\ F=m×a \\ \rm \longrightarrow a = \bigg(\dfrac{v-u}{t}\bigg) \\ a=( t v−u ) \\ Therefore \rm \longrightarrow F=m \times \\ \bigg(\dfrac{v-u}{t}\bigg)⟶F=m×( \\ t \\ v−u \\ ) \\ Here \\ F = Force \\ m = Mass \\ a = Acceleration \\ u = Initial velocity \\ v = Final velocity \\ t = Time \\ Units \\ F = Newton (N) \\ m = kilogram (kg) \\ a = metre/square second (m/s²) \\ u = metre/second (m/s) \\ v = metre/second (m/s) \\ t = second (s) \\ According to the question \\ We are asked to find the force \\ Therefore \\ We must find "F" \\ \\ Given that \\ A body of mass 500 kg started \: from rest and moves with a \\ speed of 10 m/s in 5 s \\ Hence \\ m = 500 kg \\ u = 0 m/s [starts from rest] \\ v = 10 m/s \\ t = 5 s \\ Substituting the values \\ We get \\ \rm \longrightarrow F=500 \\ \times \bigg(\dfrac{10-0}{5}\bigg) \\ \ N⟶F=500×( \\ 5 \\ 10−0 \\ ) N \\ \rm \longrightarrow F=500 \\\times\bigg(\dfrac{10}{5}\bigg) \\ \ N⟶F=500×( \\ 5 \\ 10 \\ ) N \\ \rm \longrightarrow F=100 \\ \times 10 \ N⟶F=100×10 N \\ Therefore \\ \\ \rm \longrightarrow F=1000 \\ \N⟶F=1000 N \\ Hence \\Force = 1000 N$