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4 years ago

What is the molarity of a solution made by adding 1.300 g of NaOH to a 250-mL volumetric flask and diluting to the mark?

1 answer:

3 0

To find mol with the 1.3g of NaOH divide it by the weight of NaOH.
And convert 250ml into L.
That is 0.250 ml
molarity is mol/L = 0.13 mol/L

You might be interested in

How many moles of Carbon are in 3.06 g of Carbon

natta225 [31]

Answer:

$\boxed {\boxed {\sf 0.255 \ mol \ C }}$

Explanation:

If we want to convert from grams to moles, the molar mass is used. This is the mass of 1 mole. They are found on the Periodic Table as the atomic masses, but the units are grams per mole (g/mol) instead of atomic mass units (amu).

Look up the molar mass of carbon.

Carbon (C): 12.011 g/mol

Set up a ratio using the molar mass.

$\frac {12.011 \ g \ C}{ 1 \ mol \ C}$

Since we are converting 3.06 grams to moles, we multiply by that value.

$3.06 \ g \ C*\frac {12.011 \ g \ C}{ 1 \ mol \ C}$

Flip the ratio. This way, the ratio is still equivalent, but the units of grams of carbon cancel.

$3.06 \ g \ C* \frac{1 \ mol \ C}{12.011 \ g\ C}$

$3.06 * \frac{1 \ mol \ C}{12.011 }$

$\frac {3.06}{12.011 } \ mol \ C$

$0.25476646 \ mol \ C$

The original measurement of grams (3.06) has 3 significant figures, so our answer must have the same. For the number we calculated, that is the thousandth place.

0.25476646

The 7 in the ten-thousandth place tells us to round the 4 up to a 5.

$0.255 \ mol \ C$

3.06 grams of carbon is approximately 0.255 moles of carbon.

3 0

3 years ago

How many grams of carbon are contained in one mole of<img src="https://tex.z-dn.net/?f=C_%7B3%7DH_%7B8%7D" id="TexFormula1" titl

NNADVOKAT [17]

$\quad \huge \quad \quad \boxed{ \tt \:Answer }$

$\qquad \tt \rightarrow \:36 \:\: g$

____________________________________

$\large \tt Explanation \: :$

The given compound has 3 carbon atoms, so in 1 mole of that compound, there will be 3 moles of carbon atoms.

Mass of each mole of carbon atoms = 12 g

For 3 mole carbon atoms, it will be 12 × 3 = 36 grams

Answered by : ❝ AǫᴜᴀWɪᴢ ❞

4 0

2 years ago

A 250. ML sample of 0.3M HCl is partially neutralized by the addition of 100. ML of 0.25M NaOH.Find the concentration of hydroch

Dahasolnce [82]

Answer:- 0.143 M

Solution:- HCl and NaOH reacts in 1:1 mol ratio as shown in the below reaction:

$HCl(aq)+NaOH(aq)\rightleftharpoons NaCl(aq)+H_2O(l)$

Let's calculate the initial moles of HCl and the moles of NaOH added to it:

$250.mL(\frac{1L}{1000mL})(\frac{0.3molHCl}{1L})$

= 0.075 mol HCl

$100.mL(\frac{1L}{1000mL})(\frac{0.25molNaOH}{1L})$

= 0.025 mol NaOH

Since they react in 1:1 mol ratio, 0.025 mol of NaOH will react with 0.025 moles of HCl.

Remaining moles of HCl = 0.075 - 0.025 = 0.050

Total volume of the resulting solution = 0.250 L + 0.100 L = 0.350 L

So, the concentration of HCl in the resulting solution = $\frac{0.050mol}{0.350L}$

= 0.143 M

Hence, the concentration of HCl acid in the resulting solution is 0.143 M.

6 0

4 years ago

Which statement is true about a catalyst?

kicyunya [14]

Answer:

The answer should be A. Hope this helps!

7 0

4 years ago

What is the % yield when 140.0 grams of Ethylene gas (C2H4) reacts with excess chlorine to form 280.0 grams of 1,2-Dichloro Etha

KatRina [158]

Answer:

percent yield = 56.6 %

Explanation:

Data given:

mass of Ethylene gas (C₂H₄) = 140 g

actual yield of 1,2-Dichloro Ethane (C₂H₄Cl₂)= 280 g

percent yield of 1,2-Dichloro Ethane (C₂H₄Cl₂) = ?

Reaction Given:

C₂H₄ + Cl₂ -------> C₂H₄Cl₂

Solution:

First we have to find theoretical yield.

So,

Look at the reaction

C₂H₄ + Cl₂ -----—> C₂H₄Cl₂

1 mol 1 mol

As 1 mole of C give 1 mole of CH₄

Convert moles to mass

molar mass of C₂H₄ = 2(12) + 4(1)

molar mass of C₂H₄ = 24 + 4

molar mass of C₂H₄ = 28 g/mol

molar mass of C₂H₄Cl₂ = 2(12) + 4(1) + 2(35.5)

molar mass of C₂H₄Cl₂ = 24 + 4 + 71

molar mass of C₂H₄Cl₂ = 99 g/mol

Now

C₂H₄ + Cl₂ -----—> C₂H₄Cl₂

1 mol (28 g/mol) 1 mol (99 g/mol)

28 g 99 g

28 grams of Ethylene gas (C₂H₄) produce 99 grams of C₂H₄Cl₂

if 28 g of C₂H₄ produce 99 g of C₂H₄Cl₂ so how many grams of C₂H₄Cl₂ will be produced by 140 g of C₂H₄.

Apply Unity Formula

28 g of C₂H₄ ≅ 99 g of C₂H₄Cl₂

140 g of C₂H₄≅ X of C₂H₄Cl₂

Do cross multiply

mass of C₂H₄Cl₂ = 99 g x 140 g / 28 g

mass of C₂H₄Cl₂ = 495 g

So the Theoretical yield of 1,2-Dichloro Ethane (C₂H₄Cl₂)= 495 g

Now Find the percent yield of 1,2-Dichloro Ethane (C₂H₄Cl₂)

Formula Used

percent yield = actual yield /theoretical yield x 100 %

Put value in the above formula

percent yield = 280g / 495 g x 100 %

percent yield = 56.6 %

percent yield of 1,2-Dichloro Ethane (C₂H₄Cl₂) = 56.6 %

8 0

4 years ago