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1 year ago

What is the molar concentration of a 2 liter solution containing 200 grams of glucose?

Chemistry

1 answer:

NARA [144]1 year ago

8 0

The molar concentration is 1.11M.

<h3>What is molar concentration?</h3>

The phrase "molar concentration" (also known as "molarity," "amount concentration," or "substance concentration") refers to the amount of a substance per unit volume of solution and is used to describe the concentration of a chemical species, specifically a solute, in a solution. The most frequent measure of molarity in chemistry is the number of moles per liter, denoted by the unit symbol mol/L or mol/dm3 in SI units. A solution with a concentration of 1 mol/L is referred to as 1 molar, or 1 M.

<h3>Given : </h3>

Volume of the solution = 2L

Mass of glucose given = 200g

Concentration of glucose= ?

<h3>Formula use: </h3>

Molarity = no. of moles of solute / volume of the solution (L)

Moles of solute = given mass of solute / molar mass of the solute

<h3>Solution: </h3>

No. of moles of solute( glucose ) = 200 / 180 = 1.11 moles'

Molarity = 1.11 / 2 = 0.5555 mol L ^(-1)

Therefore, the molar concentration of glucose in the solution = 0.555 mol L ^(-1)

To learn more about molar concentration :

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Identify some other substances (besides KCl) that might give a positive test for chloride upon addition of AgNO3. do you think i

nlexa [21]

Answer:

-The other substances that give a positive test with AgNO3 are other chlorides present, iodides and bromide.

-It is reasonable to exclude iodides and bromides but it is not reasonable to exclude other chlorides

Explanation:

In the qualitative determination of halogen ions, silver nitrate solution(AgNO3) is usually used. Now, various halide ions will give various colours of precipitate when mixed with with silver nitrate. For example, chlorides(Cl-) normally yield a white precipitate, bromides(Br-) normally yield a cream precipitate while iodides (I-) normally yield a yellow precipitate. Thus, all these ions or some of them may be present in the system.

With that being said, if other chlorides are present, they will also yield a white precipitate just like KCl leading to a false positive test for KCl. However, since other halogen ions yield precipitates of different colours, they don't lead to a false test for KCl. Thus, we can exclude other halides from the tendency to give us a false positive test for KCl but not other chlorides.

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2 years ago

How many hydrogen atoms are in 3 molecules of ethyl alcohol, C2H3OH

Andrew [12]

Answer: 3 hydrogen atoms

Explanation: H=hydrogen

8 0

2 years ago

What is the mass of 8.12 × 10^23 molecules of CO2 gas? (Atomic mass of carbon = 12.011 u; oxygen = 15.999 u.)

amid [387]

Answer:

$m=59.3gCO_2$

Explanation:

Hello,

In this case, the first step is to compute the molar mass of carbon dioxide as shown below, considering it has one carbon atom and two oxygen atoms:

$M=12.011g/mol+2*15.999g/mol\\\\M=44.009g/mol$

It is important to notice it is the mass in one mole of such compound. Afterwards, we need to use the Avogadro's number to compute the how many moles are in the given molecules of carbon dioxide as shown below:

$mol=8.12x10^{23}molec*\frac{1mol}{6.022x10^{23}molec} =1.35mol$

Finally, the mass by using the molar mass:

$m=1.35mol*\frac{44.009g}{1mol} \\\\m=59.3gCO_2$

Best regards.

4 0

2 years ago

if 11.74 liters of gas at STP are pumped into a basketball, how many moles of gas are in the basketball? assume the basketball w

ser-zykov [4K]

Answer:

0.52 mol

Explanation:

Using the general gas equation formula:

PV = nRT

Where;

P = pressure (atm)

V = volume (Liters)

n = number of moles (mol)

R = gas law constant (0.0821 Latm/molK)

T = temperature (K)

At STP (standard temperature and pressure), temperature of a gas is 273K, while its pressure is 1 atm

Using PV = nRT

n = PV/RT

n = (1 × 11.74) ÷ (0.0821 × 273)

n = 11.74 ÷ 22.41

n = 0.52 mol

There are 0.52 moles in the basketball

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2 years ago

A 40.0 L balloon is filled with air at sea level (1 atm @ 25 oC). It is then tied to a rock and thrown into a cold lake and it s

velikii [3]

To solve this question you need to calculate the number of the gas molecule. The calculation would be:
PV=nRT
n=PV/RT
n= 1 atm * 40 L/ (0.082 L atm mol-1K-<span>1 * 298.15K)
</span>n= 1.636 moles

The volume at bottom of the lake would be:
PV=nRT
V= nRT/P
V= (1.636 mol * 277.15K* 0.082 L atm mol-1K-1 )/ 11 atm= <span>3.38 L</span>

8 0

2 years ago