Question

4. Find the mass, in grams, of 4.21 x 1023 molecules of N2 5. How many particles are there in 2.34 g of a molecular compound with a gram molecular mass of 233 g? T2 6. Aspartame is an artificial sweetener that is 160 times sweeter than sucrose (table sugar) when iissolved in water. It is marketed by G.D. Searle as Nutra Sweet. The molecular formula of aspartame is C14H18N205 a) Calculate the gram-formula-mass of aspartame. To ed b) How many moles of molecules are in 14.3 g of aspartame? c) What is the mass in grams of 2.45 moles of aspartame? of aspartame? d) How many molecules are in 35 mg l e) How many atoms of nitrogen are in 2.3 grams of aspartame?

Answer 1

Following is the - complete Answer -&- Explanation : for the first question ( i.e. Question - 4 ), of the given: Question Set.....in....typed format...

$\Rightarrow$Answer:

Mass of  4.21 x 10²³  molecules of N₂ = 19.58 g ( grams ) ; approx.

$\Rightarrow$Explanation:

Following is the complete Explanation: for the above: Answer:

• Given:

1. Number of molecules of Nitrogen ( N₂) =  4.21 x 10²³ molecules

• ​​​​​​​Step - 1:

We know: under STP, 1 mol of N₂  : will contain =   6.022 x 10²³ molecules of N₂  ( i.e. Nitrogen ) ; i.e. the Avogadro's Number....

$\Rightarrow$ i.e. Under STP : 28.014 g ( grams ) of Nitrogen, will contain: 6.022 x 10²³ molecules : of Nitrogen ( N₂)

OR,

$\Rightarrow$ Under STP,   6.022 x 10²³ molecules : of Nitrogen ( N₂) will have the mass of : 28.014 g ( grams )

Therefore, under STP:

$\Rightarrow$4.21 x 10²³ molecules: of Nitrogen ( N₂) will have: the mass

= [ ( 28.014 g ) / ( 6.022 x 10²³ molecules) ] x ( 4.21 x 10²³ molecules )

= 19.58 g ( grams )

• Answer:

​​​​​​​Therefore: mass of : 4.21 x 10²³ molecules: of Nitrogen ( N₂) = 19.58 g ( grams ) ; approx.

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