How many grams of glucose (C6H12O6) can prepare 1750 mL of a 2.50 M solution?

787.5 grams of glucose (C₆H₁₂O₆) can prepare 1750 mL of a 2.50 M solution

Explanation:

Molarity (M) is a concentration measure that indicates the number of moles of solute that are dissolved in a given volume.

Molarity is expressed in the following way:

Molarity is expressed in units .

2.50 M  means that in 1 L of solution there are 2.5 moles of glucose. So, you apply a rule of three as follows: if in 1 L there are 2.5 moles of glucose, in 1.75 L (1750 mL, being 1000 mL = 1 L) how many moles of the compound are there?

moles=4.375

Being:

The molar mass of glucose is:

C₆H₁₂O₆= 6* 12 g/mole+ 12* 1 g/mole + 6* 16 g/mole

C₆H₁₂O₆= 180 g/mole

Then you can apply a rule of three as follows, knowing the moles in 1750 mL and the molar mass: if there are 180 g of glucose in 1 mole, how much mass is there in 4.375 moles?

mass= 787.5 g

787.5 grams of glucose (C₆H₁₂O₆) can prepare 1750 mL of a 2.50 M solution

explanation:

let the different atoms be x and y. each with four protons and same mass will be represented as follows.

mass number is number of protons + number of neutrons.

            and  

since, it is given that one has a negative charge while the other is neutral. therefore, there will be increase in number of protons and the atom will be represented as follows.

according to periodic table, mass number of beryllium is 9 and it has atomic number 4 so, the atoms will be and .  

the protons and neutrons will be located inside the nucleus whereas the electrons will be revolving around the nucleus.  

it forms an anion that has a larger radius. anions have a negative charge and are larger than the original atom.