All categories

4 years ago

What is the molarity of sodium chloride in solution that is 13.0% by mass sodium chloride and that has a density of 1.10 g/mL?

Chemistry

1 answer:

Andrej [43]4 years ago

8 0

Answer:

molarity: 2.4 M

Explanation:

1. Suppose you have 100 mL de solution:

Use density to find grams of soluton:

100 mL solution x (1.10 g / 1 mL) = 110 g solution

2. FInd the amount of sodium chloride pure:

110 g solution x(13 g NaCl / 100 g solution) = 14.3 g NaCl

3. Change grams to mol:

14.3 g NaCl x ( 1 mol NaCl/ 58.44 g NaCl) = 0.24 mol NaCl

4. Molarity:

M = mol NaCl / L solution

M = 0.24 mol NaCl / 0.1 L = 2.4 M

You might be interested in

What happens when you heat a sugar and water mixture?

a_sh-v [17]

Sugar dissolves faster in hot water than it does in cold water because hot water has more energy than cold water. When water is heated, the molecules gain energy and, thus, move faster. As they move faster, they come into contact with the sugar more often, causing it to dissolve faster.

3 0

2 years ago

Read 2 more answers

Calculate the unit cell edge length for an 85 wt% fe-15 wt% v alloy. All of the vanadium is in solid solution, and, at room temp

Lady bird [3.3K]

Answer is 0.289nm.

Explanation: The wt % of Fe and wt % of V is given for a Fe-V alloy.

wt % of Fe in Fe-V alloy = 85%

wt % of V in Fe-V alloy = 15%

We need to calculate edge length of the unit cell having bcc structure.

Using density formula,

$\rho_{ave}=\frac{Z\times M_{ave}}{a^3\times N_A}$

For calculating edge length,

$a=(\frac{Z\times M_{ave}}{\rho_{ave}\times N_A})^{1/3}$

For calculating $M_{ave}$ , we use the formula

$M_{ave}= \frac{100}{\frac{(wt\%)_{Fe}}{M_{Fe}}+\frac{(wt\%)_{V}}{M_V}}$

Similarly for calculating $(\rho)_{ave}$ , we use the formula

$\rho_{ave}= \frac{100}{\frac{(wt\%)_{Fe}}{\rho_{Fe}}+\frac{(wt\%)_{V}}{\rho_V}}$

From the periodic table, masses of the two elements can be written

$M_{Fe}= 55.85g/mol$

$M_{V}=50.941g/mol$

Specific density of both the elements are

$(\rho)_{Fe}=7.874g/cm^3\\(\rho)_{V}=6.10g/cm^3$

Putting $M_{ave}$ and $\rho_{ave}$ formula's in edge length formula, we get

$a=\left [\frac{Z\left (\frac{100}{\frac{(wt\%)_{Fe}}{M_{Fe}}+\frac{(wt\%)_{Fe}}{M_{Fe}}} \right )}{N_A\left (\frac{100}{\frac{(wt\%)_V}{\rho_V}+\frac{(wt\%)_V}{\rho_V}} \right )} \right ]^{1/3}$

$a=\left [\frac{2atoms/\text{unit cell}\left (\frac{100}{\frac{85\%}{55.85g/mol}+\frac{15\%}{50.941g/mol}} \right )}{(6.023\times10^{23}atoms/mol)\left (\frac{100}{\frac{85\%}{7.874g/cm^3}+\frac{15\%}{6.10g/cm^3}} \right )} \right ]^{1/3}$

By calculating, we get

$a=2.89\times10^{-8}cm=0.289nm$

7 0

3 years ago

Consider the reaction of A(g) + B(g) + C(g) => D(g) for which the following data were obtained:

Drupady [299]

Answer:

$r = k [A]^{2}[B]^{2}$

Explanation:

A + B + C ⟶ D

$\text{The rate law is } r = k [A]^{m}[B]^{n}[C]^{o}$

Our problem is to determine the values of m, n, and o.

We use the method of initial rates to determine the order of reaction with respect to a component.

(a) Order with respect to A

We must find a pair of experiments in which [A] changes, but [B] and C do not.

They would be Experiments 1 and 2.

[B] and [C] are constant, so only [A] is changing.

$\begin{array}{rcl}\dfrac{r_{2}}{r_{1}} & = & \dfrac{ k[A]_{2}^{m}}{ k[A]_{1}^{m}}\\\\\dfrac{2.50\times 10^{-2}}{6.25\times 10^{-3}} & = & \dfrac{0.100^{m}}{0.0500^{m}}\\\\4.00 & = & 2.00^{m}\\m & = & \mathbf{2}\\\end{array}\\\text{The reaction is 2nd order with respect to A}$

(b) Order with respect to B

We must find a pair of experiments in which [B] changes, but [A] and [C] do not. There are none.

They would be Experiments 2 and 3.

[A] and [C] are constant, so only [B] is changing.

$\begin{array}{rcl}\dfrac{r_{3}}{r_{2}} & = & \dfrac{ k[B]_{3}^{n}}{ k[B]_{2}^{n}}\\\\\dfrac{1.00\times 10^{-1}}{2.50\times 10^{-2}} & = & \dfrac{0.100^{n}}{0.0500^{n}}\\\\4.00 & = & 2.00^{n}\\n & = & \mathbf{2}\\\end{array}\\\text{The reaction is 2nd order with respect to B}$

(c) Order with respect to C

We must find a pair of experiments in which [C] changes, but [A] and [B] do not.

They would be Experiments 1 and 4.

[A] and [B] are constant, so only [C] is changing.

$\begin{array}{rcl}\dfrac{r_{4}}{r_{1}} & = & \dfrac{ k[C]_{4}^{o}}{ k[C]_{1}^{o}}\\\\\dfrac{6.25\times 10^{-3}}{6.25\times 10^{-3}} & = & \dfrac{0.0200^{o}}{0.0100^{o}}\\\\1.00 & = & 2.00^{o}\\o & = & \mathbf{0}\\\end{array}\\\text{The reaction is zero order with respect to C.}\\\text{The rate law is } r = k [A]^{2}[B]^{2}$

5 0

3 years ago

Which organic compound is isometric with at least one aldehyde?

azamat

D. Ketone :::::::::::

4 0

4 years ago

How do you number 1 and 2

Georgia [21]

1. You can use the formula V1/ T1 = V2/T2
Solve for V2 give than V1= 10.0 L, T1= 25 C= 298 K, T2= 10 C= 283 K.

(10.0L/298K)= V2/283K------> V2= 9.50 L.

I cant see the answer there.. weird!

2. you can use the formula P1V1= P2V2
solve for V2 given than P1= atm, V1= 2.0 L, P2= 300 atm.

(1atm)x(2.0L)= (300atm)V2----->0.00667 atm.

so answer is A.

8 0

3 years ago