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2 years ago

What is the molarity of a 50.0 ml solution containing 10.0 grams of table sugar (C12H22O11)?

Chemistry

1 answer:

Lostsunrise [7]2 years ago

6 0

Answer: The molarity of solution is 0.584 M

Explanation:

Molarity of a solution is defined as the number of moles of solute dissolved per liter of the solution.

$Molarity=\frac{n\times 1000}{V_s}$

where,

n = moles of solute

$V_s$ = volume of solution in ml

mass of solute (sugar) = $\frac{\text {given mass}}{\text {Molar Mass}}=\frac{10.0g}{342g/mol}=0.0292$

Now put all the given values in the formula of molality, we get

$Molarity=\frac{0.0292\times 1000}{50.0}=0.584mole/L$

Therefore, the molarity of solution is 0.584 M

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A 30.0 g sample of a metal was heated in a hot water bath to 80°c. it was then quickly transferred to a coffee-cup calorimeter.

Annette [7]

when the metal is lost heat and the calorimeter of water is gained the heat

and when the heat lost = the heat gained so,

(M*C*ΔT)m = (M*C*ΔT)w

when Mm= mass of the metal = 30 g

Δ Tm = (80-25) = 55 °C

and Mw = mass of water = 100 g

Cw is the specific heat of water = 4.181 J/g.°C

ΔTw = (25-20) = 5 °C

so by substitution:

∴ 30* Cm*55 = 100 * 4.181 * 5

∴Cm (specific heat of metal) = (100*4.181*5)/(30*55)

∴C of metal = 1.267 J/g.°C

3 0

2 years ago

Read 2 more answers

A certain element X has four isotopes. 4.350% of X has a mass of 49.94605 amu. 83.79% of X has a mass of 51.94051 amu. 9.500% of

sesenic [268]

Answer: The average atomic mass of the element X is 51.99592 amu

Explanation:

Mass of isotope 1 = 49.94605 amu

% abundance of isotope 1 = 4.350% = $\frac{4.350}{100}=0.0435$

Mass of isotope 2 = 51.94051 amu.

% abundance of isotope 2 = 83.79% = $\frac{83.79}{100}=0.8379$

Mass of isotope 3 = 52.94065 amu.

% abundance of isotope 2 = 9.500% = $\frac{9.500}{100}=0.095$

Mass of isotope 4 = 53.93888 amu.

% abundance of isotope 2 = 2.360% = $\frac{2.360}{100}=0.0236$

Formula used for average atomic mass of an element :

$\text{ Average atomic mass of an element}=\sum(\text{atomic mass of an isotopes}\times {{\text { fractional abundance}})$

$A=(49.94605\times 0.0435)+(51.94051 \times 0.8379)+ (52.94065\times 0.095)+(53.93888\times 0.0236)$

$A=51.99592amu$

Therefore, the average atomic mass of the element X is 51.99592 amu

5 0

2 years ago

Solving applied density problems Mr. Auric Goldfinger, criminal mastermind, intends to smuggle several tons of gold across Inter

Margarita [4]

Answer:

thickness = 0.29 cm

Explanation:

In order to make fake iron ball [made of gold] we should get mass of fake ball

should be equal to that of Iron ball.So for that we should calculate

volume of iron ball using diameter given ;formula is 4/3 pi r^3

given d= 6 cm;so radius r= 6/2 = 3 cm

then volume of Iron ball = 4/3 *3.14* 3^3 = 113.04 cm^3

So mass of iron ball = volume x density = 113.04 * 5.15 g/cm^3 = 582.156 g

This must be the mass of gold ball ;now calculate volume of gold ball using its

density

volume of gold ball = mass of gold ball/density of gold ball = 582.156 g/19.3 g/cm^3

= 30.1635 cm^3

Now this must be the volume of hollow sphere whose outer radius R = 3cm

and inner radius r= ??

Volume of hollow ball = 4/3pi[R3-r^3]

30.1635 cm^3 = 4/3 pi [3^3-r^3]

30.1635* 3/4*3.14 = 27 - r^3

7.2046 = 27- r^3

r^3 = 19.7954

r= 2.7051 cm

So the thick ness = outer radius- inner radius = 3 - 2.7051 = 0.2949 cm

rounding to 2 significant figures

we get thickness = 0.29 cm

8 0

2 years ago

Sulfur and oxygen react to produce sulfur trioxide. In a particular experiment, 7.9 grams of SO3 are produced by the reaction of

shutvik [7]

Answer:

79%

Explanation:

1) Balanced chemical equation:

2S + 3O₂ → 2SO₃

2) Mole ratio:

2 mol S : 3 mol O₂ : 2 mol SO₃

3) Limiting reactant:

Number of moles of O₂

n = 6.0 g / 32.0 g/mol = 0.1875 mol O₂

Number of moles of S:

n = 7.0 g / 32.065 g/mol = 0.2183 mol S

Ratios:

Actual ratio: 0.1875 mol O₂ / 0.2183 mol S =0.859

Theoretical ratio: 3 mol O₂ / 2 mol S = 1.5

Since there is a smaller proportion of O₂ (0.859) than the theoretical ratio (1.5), O₂ will be used before all S be consumed, and O₂ is the limiting reactant.

4) Calcuate theoretical yield (using the limiting reactant):

0.1875 mol O₂ / x = 3 mol O₂ / 2 mol SO₃

x = 0.1875 × 2 / 3 mol SO₃ = 0.125 mol SO₃

5) Yield in grams:

mass = number of moles × molar mass = 0.125 mol × 80.06 g/mol = 10.0 g

6) Percent yield:

Percent yield, % = (actual yield / theoretical yield) × 100

% = (7.9 g / 10.0 g) × 100 = 79%

6 0

2 years ago

A 0.0035 M aqueous solution of a particular compound has pH = 2.46. The compound is (A) a weak base (B) a weak acid (C) a strong

slava [35]

Answer:

(a) A strong acid

Explanation:

We have given the pH of the solution is 2.46

pH=2.46

So the concentration of $H^+=10^{-pH}=10^{-2.46}=0.00346$

solution having H+ concentration more than $H^+=10^{-7}$ is acidic

Since in the given solution, H+ concentration is 0.00346 M which is more than 10^{-7}[/tex] so this is an acidic solution

Note-The concentration of $H^+$ decide the behavior of the solution that is, it is acidic or basic

7 0

2 years ago